Ghasem D. Najafpour. Biochemical Engineering and Biotechnology

Conversion

(E.3.11)

For t ⫽ 10 min, X

A

⫽ 1 – e

–0.2565

⫽ 0.226 or 22.6%.

For t ⫽ 20 min, X

A

⫽ 1 – e

–0.02565(20)

⫽ 0.401 or 40.1%.

For t ⫽ 30 min, X

A

⫽ 1 – e

–0.02565(30)

⫽ 0.5367 or 53.67%.

For t ⫽ 60 min, X

A

⫽ 1 – e

–0.02565(60)

⫽ 0.785 or 78.5%.

Example 4

In a complex enzyme reaction, multiple substrate–enzyme complexes are formed. Assume

the following reaction mechanisms are taking place in three consecutive stages:

(E.4.1)

Develop a suitable rate expression using the Michaelis–Menten rate equation and the quasi-

steady-state approximations for the intermediate complexes formed.

Solution

Equilibrium dissociation constants are defined as

(E.4.2)

(E.4.3)

(E.4.4)

(E.4.5)

(E.4.6)

(E.4.7)

v

k

K

ES

k

KK

ES⫽⫽

5

2

1

5

12

[] [][]

[]

[][]

ES

K

1

[]

ES

K

2

1

2

⫽

y ⫽ kES

52

[]

K

k

ES

2

4

3

1

2

⫽⫽

[]

K

k

ES

1

2

11

⫽⫽

[][]

[]

S E ES ES P E

KK

k

⫹⫹

12

5

12

¨Ææ¨Æææ æ Ææ

X

S

A

o

t

⫽⫺ ⫽⫺

⫺

11

0 02565

e

.

114 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY

Ch005.qxd 10/27/2006 10:44 AM Page 114

The total enzyme concentration is:

(E.4.8)

(E.4.9)

(E.4.10)

(E.4.11)

The final rate expression is:

(E.4.12)

Example 5

Pesticide inhibition on an active enzyme has been reported, which caused enzyme activi-

ties to reduce. The collected data with and without inhibition are presented in Table E.5.1.

Determine the rate model with and without inhibitor (see Table E.5.1). Also define the type

of inhibition.

y =

Ê

Ë

Á

ˆ

¯

˜

Ê

Ë

Á

ˆ

¯

˜

ek

KK

S

K

KK

S

ke S

KK K

oo5

12

2

12

5

12 2

1

⫻

⫹

⫽

⫹⫹

[]

()

[[]

[]

S

ke

K

S

KK

K

S

o

⫽

⫹

5

2

12

2

1

Ê

Ë

Á

ˆ

¯

˜

Ê

Ë

Á

ˆ

¯

˜

y

max

⫽

⫹

ke

K

o5

2

1

Kk

KK

K

obs apparent

⫽⫽

⫹

12

2

1

E

e

K

KK

S

o

⫽

⫹

1

2

12

Ê

Ë

Á

ˆ

¯

˜

[]

E E ES ES E

ES

K

ES

KK

ES

KKK

012

112 112

1

11

⫽⫹ ⫹ ⫹⫹ ⫹ ⫽ ⫹ ⫹

[][] [][]

È

Î

Í

˘

˚

˙

Ê

Ë

Á

ˆ

¯¯

˜

GROWTH KINETICS 115

TABLE E.5.1. Substrate concentration and enzymatic rate calculation

with and without inhibition

S, mol⭈l

⫺1

y (no inhibitor) y

*

(inhibitor)

mol⭈l

⫺1

⭈min

⫺1

⫻ 10

6

mol⭈l

⫺1

⭈min

⫺1

⫻10

6

3.30 ⫻ 10

⫺4

56 37

5.00 ⫻ 10

⫺4

71 47

6.70 ⫻ 10

⫺4

88 61

1.65 ⫻ 10

⫺3

129 103

2.21 ⫻ 10

⫺3

149 125

Ch005.qxd 10/27/2006 10:44 AM Page 115

Solution

Plot both sets of data as a Lineweaver–Burk plot for competitive inhibition (see Fig. E.5.1):

(E.5.1)

The mechanism of the enzyme with substrate in the present of inhibitor is:

(E.5.2)

(E.5.3)

(E.5.4)

(E.5.5)

Without inhibition:

(E.5.6)

(E.5.7)

(E.5.8)

(E.5.9)

(E.5.10)

k

i

0 786

10

0 786

173 10

5

3

..

.⫽⫽⫻

⫺

m

i

k

i

⫽ 0 786.

1 5 10 8 4 10 1

34

..⫻⫽⫻ ⫹

⫺⫺

i

k

i

Ê

Ë

Á

ˆ

¯

˜

KK

i

k

ss

i

app

⫽⫹1

Ê

Ë

Á

ˆ

¯

˜

y ⫽

⫹

⫽

⫹⫻

⫺

ke S

SK

ke S

S

o

s

o

84 10

3

.

e E ES EI

o

⫽⫹ ⫹[][]

ES E P ⫹

EI EI⫹ 

ES ES

K

S

⫹ ¨Æææ

y ⫽

⫹⫹

ke S

SK

i

k

o

s

i

1

Ê

Ë

Á

ˆ

¯

˜

116 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY

TABLE E.5.2. Inverse substrate concentration and inverse enzymatic rate calculation with and

without inhibition

1/S,l⭈mol

⫺1

y (no inhibitor) 1/yy

*

(inhibitor) 1/y

*

,

mol⭈l

⫺1

⭈min

⫺1

⫻ 10

6

mol⭈l

⫺1

⭈min

⫺1

⫻ 10

6

l⭈min⭈mol

⫺1

3030 56 17857 37 27027

2000 71 14085 47 21277

1492 88 11363 61 16393

606 129 7752 103 9709

452 149 6711 125 8000

Ch005.qxd 10/27/2006 10:44 AM Page 116

Plug in a number

(E.5.11)

(E.5.12)

(E.5.13)

The reading from the plot for the rate without inhibition:

For the data plotted with inhibition:

Slope

()

.

18 5 10

3100

42

3

⫺⫻

⫽

k

app

⫽⫻

⫺

15 10

3

.

Slope

()

.

27 5 10

3000

73

3

⫺⫻

⫽

i

k

i

⫽ 049.

61 10

12 10 67 10

6 7 10 8 4 10 1

6

34

43

⫻⫽

⫻⫻⫻

⫻⫹⫻ ⫹

⫺

⫺⫺

..

i

k

i

Ê

Ë

Á

ˆ

¯

˜

ke

o

⫽⫻

⫺⫺

1.2 10 min

31

88 10

67 10

67 10 84 10

6

4

43

⫻⫽

⫻⫻

⫻⫹⫻

⫺

⫺⫺

ke

o

.

..

GROWTH KINETICS 117

1/S, l/mol

-1000 0 1000 2000 3000

1/ν, l.min.mol

-1

0

5000

10000

15000

20000

25000

30000

35000

Without inhibition

with inhibition

FIG. E 5.1. Competitive inhibition based on the Lineweaver-Burk model.

Ch005.qxd 10/27/2006 10:44 AM Page 117

Based on the graphical presentation, this is a competitive inhibition.

Example 6

The respiratory quotient (RQ) is often used to estimate metabolic stoichiometry. Using

quasi-steady-state and by definition of RQ, develop a system of two linear equations with

two unknowns by solving a matrix under the following conditions: the coefficient of the

matrix with yeast growth (g ⫽ 4.14), ammonia (g

N

⫽ 0) and glucose (g

S

⫽ 4.0), where the

evolution of CO

2

and biosynthesis are very small (s ⫽ 0.095). Calculate the stoichiometric

coefficient for RQ ⫽ 1.0 for the above biological processes:

Solution

At quasi-steady-state condition:

(E.6.1)

(E.6.2)

(E.6.3)

The mathematical solution for the above system is set as given by the following matrix:

(E.6.4)

At singular point, let

(E.6.5)

v

RQ

S

2

1

0

⫺

⫽

()

È

Î

Í

˘

˚

˙

v

RQ

a

b

vv

n

v

S

BS N

2

1

2

1

⫺

⫽

⫺⫺

⫺

()

È

Î

Í

˘

˚

˙

È

Î

Í

˘

˚

˙

+

È

Î

Í

˘

˚

˙

s

˙˙

y

yy y

s

BS N

a

b

n

abRQ

2

1

2

1⫺⫽ ⫺

⫺⫽

()

+-

È

Î

Í

˘

˚

˙

-

Ï

Ì

Ô

Ó

Ô

d

RQ

a

b

⫽⫽

⫹

moles of CO

moles of O

2

s

d[ ]

d

NADH H

t

ab

n

s

ss N

⫹

⫽⫻⫺⫺ ⫺ ⫺⫻ ⫽

⫹

y

yy y

2

1

2

10()+

È

Î

Í

˘

˚

˙

s

d

y

max

⫽⫻

⫺

210

4

mol/min

K

s

⫽⫻

⫺

84 10.

4

mol/L

K

s

y

max

.⫽ 42

118 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY

Ch005.qxd 10/27/2006 10:44 AM Page 118

(E.6.6)

(E.6.7)

For anaerobic fermentation, y

S

⫽ 4 then RQ ⫽ 1.0

(E.6.8)

(E.6.9)

Increasing RQ with very small changes causes the constants a and b to change in the reaction.

Example 7

For single- and multiple-substrate kinetics, single-substrate glucose, 30g⭈l

⫺1

and dual

substrates glucose and lactose with each carbohydrate at a concentration of 15 g⭈l

⫺1

or total

22

1105

012

0 095

052 058

⫺

⫽

⫺

⫽⫽

.

., .

È

Î

Í

˘

˚

˙

È

Î

Í

˘

˚

˙

È

Î

Í

˘

˚

˙

a

b

ab

22

1105

1

2

4 14 4 1 0 095 0

0 095

⫺

⫽

⫺⫹ ⫺

⫺

.

[. ( . ) ]

.

È

Î

Í

˘

˚

˙

È

Î

Í

˘

˚

˙

È

Î

Í

˘

˚

a

b

˙˙

˙

È

Î

Í

˘

˚

˙

⫽

⫺

⫽⫽

012

0 095

064 07

.

., .ab

RQ

S

⫽

4

y

⫺⫺⫺⫽() ()()RQ

S

y

2

21 0

Ê

Ë

Á

ˆ

¯

˜

GROWTH KINETICS 119

TABLE E.7.1. Optical density for single and double substrates

Time, h OD, single OD, double Time, h OD, single OD, double

0.0 0.05 0.05 4.5 0.89 0.44

0.5 0.08 0.06 5.0 1.05 0.49

1.0 0.11 0.07 5.5 1.05 0.51

1.5 0.14 0.08 6.0 1.06 0.53

2.0 0.20 0.10 6.5 1.06 0.61

2.5 0.27 0.14 7.0 1.08 0.85

3.0 0.38 0.19 7.5 1.08 1.04

3.5 0.50 0.27 8.0 1.08 1.04

4.0 0.71 0.33 — — —

t

lag

⫽ 1 h for dual substrate.

t

lag

⫽ 0 for single substrate.

Ch005.qxd 10/27/2006 10:44 AM Page 119

carbohydrate concentration of 30 g⭈l

⫺1

were used in sterilised media (Table E.7.1). The cell

optical density (OD) was measured at wavelength 420 nm. The following data for single

and double substrates were obtained. If OD is linear in cell density with a value of 0.175

equal to 0.1 mg cell dry weight per millilitre, evaluate the specific growth rate, lag phase

and yield of biomass.

Solution

(E.7.1)

(E.7.2)

For t ⫽ 5h:

(E.7.3)

For t ⫽ 8h:

Therefore, yield is:

The shape of dual substrates with the same total concentration of sugar and lactic acid

shows two stages like stepwise utilisation of dual substrates has taken place in the cells.

Y ⫽⫽

(. )

.

104

01

0 175

30

002

È

Î

Í

˘

˚

˙

m

h⫽

⫺

057

1

. Single substrate

m

⫽

ln

.

10

005

8

Ê

Ë

Á

ˆ

¯

˜

È

Î

Í

˘

˚

˙

m

⫽

⫺

05. h dual substrate (Fig. E.7.1)

1

N() .

.

0005

01

0 175

005

0 175

⫽⫽

g dry cell

Ê

Ë

Á

ˆ

¯

˜

01

0 175

104

005

0 175

5

.

(. ) ( )

.

Ê

Ë

Á

ˆ

¯

˜

Ê

Ë

Á

ˆ

¯

˜

⫽⫽NS

m

e

m

Nt Noe

m

t

()⫽

m

1

n

t

d

⫽ m

120 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY

Ch005.qxd 10/27/2006 10:44 AM Page 120

Example 8

Monod kinetics are considered in a CSTR with an organism growing with an initial sub-

strate concentration of 50 g⭈l

⫺1

and kinetic parameters of K

s

⫽ 2g⭈l

⫺1

and m

max

⫽ 0.5 h

⫺1

.

(a) What would be the maximum dilution rate for 100% yield of biomass with maximum

rate? (b) If the same dilution is used, what would be number of CSTRs in series?

Solution

(a)

(E.8.1)

(E.8.2)

(b) For the first tank, use (5.6.7.5), obtained from the steady-state condition, then solved for

substrate concentration profile. The outlet substrate concentration was calculated as:

(E.8.3)

S

DK

D

s

1

0 402 2

05 0402

⫽

⫺

⫽

⫺

⫽

⫺

m

max

(. )()

..

8.2 g l

1

⭈

⫽⫻⫺

⫹

⫽

⫺

(. )05 1

2

250

1

h

0.402 h

1

È

Î

Í

˘

˚

˙

D

K

KS

s

so

max max

⫽⫺

⫹

m 1

È

Î

Í

˘

˚

˙

GROWTH KINETICS 121

Time, h

0246810

Optical density (OD) at 420 nm

0.0

0.3

0.6

0.9

1.2

Cell dry weight, mg/l

0.0

0.3

0.6

0.9

1.2

Glucose, 30 g/l

Glucose, 15 g/l

Lactose, 15 g/l

Dual substrate has about the same

pattem as the growth curves are shown

FIG. E.7.1. Single and double substrate growth model.

Ch005.qxd 10/27/2006 10:44 AM Page 121

Biomass concentration is obtained from definition as substrate is utilised; it is converted to

cells or biomass:

(E.8.4)

Now substitute the outlet concentration of substrate, resulting in the following equation:

(E.8.5)

For the subsequent tanks, use the steady-state mass balance on substrate:

(E.8.6)

(E.8.7)

(E.8.8)

(E.8.9)

(E.8.10)

Solve for substrate concentration:

(E.8.11)

(E.8.12)

(E.8.13)

0 5 22 51 6 6 0

2

...SS⫹⫺⫽

( . )( )( . ) . ( )[ ( . )]0 402 1 8 2 0 5 8 2

221 2

⫺⫽ ⫺SSXYS+

DY S S S K S X Y S S

S

()( )( ) [ ( )]

max122 21 12

⫺⫹⫽ ⫹⫺m

DS S

SX YS S

YS K

s

()

[()]

()

max

12

21 1 2

2

0⫺⫺

⫹

⫽

m -

XXYSS

21 12

⫽⫹ ⫺()

XXYSS

21 12

⫺⫽ ⫺()

DS S

SX

YS K

s

()

max

12

22

2

0⫺⫺

⫹

⫽

m

DS S

Y

X

XS

()

/

feed

⫺⫺ ⫽

1

0m

XYS

DK

D

X

s

10

1

0 402 2

05 0402

⫽⫺

⫺

⫽⫺

⫺

⫽

m

max

(

(. )()

..

È

Î

Í

˘

˚

˙

È

Î

Í

˘

˚

˙

1) 50 441 8.gl

1

⭈

⫺

XY S S

XS

⫽⫺

/ in out

()

122 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY

Ch005.qxd 10/27/2006 10:44 AM Page 122

Solving the quadratic equation, the meaningful substrate concentration is:

S

2

⫽ 0.29 g⭈l

⫺1

N ⫽ two tanks

Example 9

For the design of a CSTR with inhibition, consider the following rate is valid for a CSTR

as a fermentation vessel.

Given the following data:

Initial substrate concentration, S

o

⫽ 10 g⭈l

⫺1

Inlet cell concentration for sterile media, X

o

⫽ 0

The rate constant, K

S

⫽ 1g⭈l

⫺1

The inhibitory constant, K

i

⫽ 0.01 g⭈l

⫺1

The inhibitor concentration, [I] ⫽ 0.05 g⭈l

⫺1

The m

max

⫽ 0.5 h

⫺1

and Yield of biomass on substrate Y

X/S

⫽ 0.1 g cells/g substrate.

Solution

The rate model with inhibition is given by (5.7.3.5).

(E.9.1)

(E.9.2)

X

0

⫽ 0 for sterile feed

(E.9.3)

Rearrangement of the rate model and the equation is solved for substrate S:

(E.9.4)

With defined values of inhibition concentration, rate constant and given dilution rate, the sub-

strate concentration is calculated based on (E.9.4). The tabulated data are given in Table E.9.1.

S

DK DK

i

K

D

ss

i

⫽

⫹

⫺

Ê

Ë

Á

ˆ

¯

˜

m

max

XY S S

XS

⫽⫺

/

()

0

D

S

SK

i

K

s

i

⫽⫽

⫹⫹

m

max

1

Ê

Ë

Á

ˆ

¯

˜

r

SX

KS

i

K

X

s

i

s

⫽

⫹⫹

m

max

GROWTH KINETICS 123

Ch005.qxd 10/27/2006 10:44 AM Page 123

Ghasem D. Najafpour. Biochemical Engineering and Biotechnology

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