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KC31003 Bioprocess Principles Group 2 Group Member: TAI WEI KEONG (BK09110063) SEAH YONG UNG (BK09110144) RUSSELL ROSS SULAU (BK09110156) SCOTT BIONDI R. VALINTINUS (BK09110151) CLARICE V. BINJINOL (BK09110005) TAN YIT ZEN (BK09110015)

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KC31003 Bioprocess Principles

Group 2

Group Member:

TAI WEI KEONG (BK09110063)

SEAH YONG UNG (BK09110144)

RUSSELL ROSS SULAU (BK09110156)

SCOTT BIONDI R. VALINTINUS (BK09110151)

CLARICE V. BINJINOL (BK09110005)

TAN YIT ZEN (BK09110015)

The growth rate of E. coli be expressed by Monod kinetics with the parameters of µmax = 0.935 hr-

1 and Ks = 0.71 g/L (Monod, 1949). Assume that the cell yield YX/S is 0.6 g dry cells per g substrate. If CX0 is 1 g/L and CS1= 10 g/L when the cells start to grow exponentially, at t0 = 0, show how ln Cx, Cx, Cs, dlnCx /dt, and dCx /dt change with respect to time.

Question 1

Given data

• µmax = 0.935 hr-1

• Ks = 0.71 g/L

• YX/S = 0.6 g

• CX0 = 1 g/L

• Cs1= 10 g/L

Rate of microbial growth is characterized by the

net specific growth rate

Net specific gross specific rate of loss

growth rate growth rate of cell mass

0, cells start to grow exponentially at

Equation (1) Equation (2)

Equation (3)

We assume that a single chemical species, S is a

growth –rate limiting. This kinetics can be

described by the Monod equation.

Equation (4)

Equation (5) Equation (6)

Based on our question, we assume

Equation (7)

Equation (8) Equation (9)

Equation (10)

SAMPLE CALCULATION

• Cx increased at a step size of 0.1.

• Cs data is calculated by using following equation:

• The value of t is calculated from the integration with the values of Cs and Cx for each iteration made is calculated based on the previous 2 assumptions that made.

Equation (11)

Cx ln Cx Cs t dlnCx/dt dCx/dt 1.0 0.0000 10.0000 0.0000 0.8720 0.9149

1.1 0.0953 9.8333 0.1093 0.8699 0.9998

1.2 0.1823 9.6667 0.2093 0.8676 1.0839

1.3 0.2624 9.5000 0.3016 0.8651 1.1674

1.4 0.3365 9.3333 0.3872 0.8624 1.2500

1.5 0.4055 9.1667 0.4672 0.8595 1.3317

1.6 0.4700 9.0000 0.5423 0.8563 1.4124

1.7 0.5306 8.8333 0.6131 0.8529 1.4922

: : : : : :

6.4 1.8563 1.0000 3.3949 0.0496 0.3199

6.5 1.8718 0.8333 3.7076 0.0332 0.2173

6.6 1.8871 0.6667 4.1677 0.0199 0.1324

6.7 1.9021 0.5000 4.9231 0.0099 0.0669

6.8 1.9169 0.3333 6.4171 0.0033 0.0225

6.9 1.9315 0.1667 10.8661 - -

0

2

4

6

8

10

12

0 2 4 6 8 10 12

Cx

and

Cs

time

Cx and Cs versus time

Cx

CS

0.0100

0.1000

1.0000

10.0000

0.0000 2.0000 4.0000 6.0000 8.0000 10.0000 12.0000

ln C

x

Time

lnCx versus time

0.0010

0.0100

0.1000

1.0000

0.0000 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000

dln

Cx/

dt

Time

dlnCx/dt versus time

0

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5 6 7

dC

x/d

t

time

dCx/dt versus time

Question 2

Immobilized cell or enzyme technology for formation of biocatalyst has been valuable tool in many bioprocesses. Among the technologies, entrapment of biological materials within calcium alginate gel beads is the most commonly used.

a) Explain why calcium alginate has been popular as immobilization material. Suggest alternative materials suitable for immobilization.

Reasons:

1) The hydrogel formation occurs under extremely mild conditions

2) Does not depend on the formation of more permanent covalent bonds between polymer chain.

3) It is biocompatible & has good mechanical strength for many different applications.

Alternative material

1) Polyacrylamide -Non-ionic -Properties of the enzymes are only minimally modified in the presence of the gel matrix. -Diffusion of the charged substrate & products is not affected.

2) Chitosan

- Obtained from alkaline hydrolysis of chitin. - It is very biocompatible -Has been used in many applications including drug delivery system. - Disadvantage:

Limited solubility in water

Require dilute acidic solution for dissolution.

b) Describe at least two purposes for using immobilization system in such fashion and give example on how the each purposed is serve.

Purposes 1)Cells last longer -Ordinary cells (only 1-2days) but immobilized cells

can last for 30days. • How? a) cells suspended into solution of sodium alginate.

b)Added to a calcium chloride solution.

c)Packed into column & a solution to be converted is fed into the top of the column

d)Allowed to flow through the bed of beads contain immobilized biocatalyst in the cells.

e)Conversion takes place & product comes out t the bottom.

-Example: immobilized yeast cells

2. Retain the protein while allowing penetration of substrate

-Example: calcium alginate

-Widely used due to the mild condition

-Result in minimum denaturation of the biocatalyst.

-Can be formed in extremely mild conditions which ensure that enzyme activity yields over 80% can be achieved.

-Entrapment in insoluble calcium alginate gel:rapid,nontoxic,inexpensive,& versatile methods.

c) Describe at least three methods for large-scale production of calcium alginate beads and state the pros and cons of each method.

Method I

Preparation of Calcium Alginate Microgel Beads in an Electrodispersion Reactor Using an Internal Source of Calcium

Carbonate Nanoparticles

• Pulsed electric fields are utilized to atomize aqueous mixtures of sodium alginate and CaCO3 nanoparticles (dispersed phase) from a nozzle into an immiscible, insulating second liquid (continuous phase) containing a soluble organic acid.

Advantages • Alginate molecules are stabilized by the

electrostatic force in a gel

• The coalescence of the emulsified drops containing sodium alginate and CaCl2 is capable of producing micrometer-sized calcium alginate (Ca-alginate) gel beads.

• The scale-up potential of this process is almost unlimited.

• The technique is suitable for biological or food applications.

• The characteristic highly porous structure of Ca-alginate particles

Disadvantages • The random process of droplet

coagulation vary widely in size and shape

• This processing involves harsh conditions that can denature biological compounds.

• The difficulty in using dispersion/external gelation techniques with ionic polysaccharide is that the calcium source (CaC12) is insoluble in the oil phase.

Method II

Production of calcium alginate beads by using Sound Wave Induced

Vibration Method

• A new process for the immobilization of calcium alginate, which employs a vibration of membrane induced by sound waves from a horn-type speaker, is developed to produce uniformly-sized beads of predictable diameter.

• Frequency of sound wave and production rate are varied to find the optimal control ranges for the bead production.

• The controllable bead size range is 1.50-3.50 mm

• The experimental production rates are 0-12 dm3 h-1 by a single nozzle.

Advantages • Alternative to conventional continuous

• Low apparatus cost

• More easier to set up compare to other methods

• Better control of bead size and shape

Disadvantages • Speed of sound varies with temperature

• Difficult to accurately measure the exact arrival time of the wave front at different sensors

Method III

Production of alginate beads by emulsification/internal gelation

• Alginate microspheres were produced by emulsification/internal gelation of alginate sol dispersed within vegetable oil.

• Gelification was initiated within the alginate solution by a reduction in pH (7.5 to 6.5), releasing calcium from an insoluble complex.

• Smooth, spherical beads with the narrowest size dispersion were obtained when using low-guluronic-acid and low-viscosity alginate and a carbonate complex as the calcium vector.

• Diameters ranging from 50 gm to 1000 tm were obtained with standard deviations ranging from 35% to 45% of the mean.

Advantages • The resultant broad size distribution.

Producing small-diameter alginate beads (200-1000 pm), potentially on a large scale (m3/h).

• Enhanced rates of bioconversion and to avoid bead rupture resulting from the accumulated fermentation gas. Ionic polysaccharide is that the calcium source.

• By controlling the conditions under which the water-in-oil dispersion is produced, the bead size can be controlled from an few micrometers to millimetres in diameter.

Disadvantages • Its cost is high in processing.

• Emulsion is unstable.

• The emulsion tends to separate into a watery layer and an oily layer.

d) Describe one industrial application for each immobilized cell and enzyme system with the aid of PID diagram.

Application of immobilized cell in Beer brewing industry

• Main advantages of using immobilised cells for production of beer are:

1- Enhanced fermentation productivity due to higher biomass densities

2- Improved cell stability

3- Easier implementation of continuous operation

4- Improved operational control and flexibility

5- Facilitated cell recovery and reuse

6- Simplified downstream processing.

(Source: Shreve, R. N., & T.Austin, G. (1984). Shreve's Chemical Process Industries. New York: McGraw-Hill.)

• Immobilisation cell system is applied in the fermenter.

• Kirin 3 stages bioreactor system.

(Source: Kirin's Three-Stage Bioreactor. (n.d.). Retrieved May 10, 2012, from SpringerImages: http://www.springerimages.com/Images/Chemistry/1-10.1007_s11947-010-0435-0-10)

Application of Immobilized Cell in Bioreactor

• Enzyme-immobilized membrane bioreactors (EMBR) is a combination of product separation process with enzymatic catalysis in continuous processes.

• The selective membrane are use to separate the enzymes from the reaction products. EMBR have been widely used in the field of fine chemicals synthesis.

(Source: Fang, Y., Huang, X.-j., Chen, P.-C., & Xu, Z.-K. (2011). Polymer materials for enzyme immobilization and their application in bioreactors.)

e)Describe the problems that envisaged when using immobilized cell/enzyme technology.

1)Immobilized enzymes - Often adversely affects the stability and activity of

the enzymes. - Used for a prolonged period of operation - Present large problems in diffusion of the

substrate to reach the enzyme. - Another problem: Leakage of enzyme Based on the physical occlusion of enzyme molecules within an enclosed gel structure diffusion of enzyme molecules to the surrounding medium is limited. Any use of enzymes as catalysts requires consideration of their operational life

2) Immobilized cell

- May cause extra diffusion limitations

- Main problem: mass transfer resistance imposed by the fact that the substrate

has to diffuse to the reaction site and inhibitory or toxic products must be removed to the environment.

- Depend on the relative rates of bioconversion and diffusion.

Question 3

• A by-product of fermentation process has shown to be inhibitory to cell growth. From the data below: – A) Obtain the kinetic parameters µm, ks.

– B) Determine the type of inhibition by the inhibitor. Justify your answer and what do you think of the data accuracy.

– C) Determine inhibition constant, kp. What can u do to improve the reliability of the value?

By (g/L) Glucose (g/L)

Specific growth

rate µ (hr-1)

By (g/L) Glucose (g/L)

Specific growth

rate µ (hr-1)

0.00 3.00 0.15 25.00 3.00 0.07

4.00 0.17 4.00 0.07

5.00 0.19 5.00 0.09

8.00 0.25 8.00 0.10

12.00 0.26 12.00 0.15

16.00 0.29 20.00 0.18

20.00 0.41

32.00

3.00 0.05

20.00 3.00 0.08 4.00 0.05

4.00 0.10 5.00 0.07

5.00 0.11 8.00 0.07

6.00 0.11 12.00 0.09

10.00 0.17 20.00 0.11

20.00 0.23

Solution 3a

• Table shows the cell growth with same substrate concentration parameter at the

• Conditions => different by-product concentration which caused inhibition to the cell growth

• By-product concentration increased from 0 to 32g/L.

• Double reciprocal or Lineweaver-Burke (LWB) plot is used to obtain ks and µmax.

Calculated value for 1/[s] and 1/[µ] By-product

(g/L) Glucose (g/L) 1/[S]

Specific growth

rate µ (hr-1) 1/µ

0.00 3.00 0.3333 0.15 6.6667

4.00 0.2500 0.17 5.8824

5.00 0.2000 0.19 5.2632

8.00 0.1250 0.25 4.0000

12.00 0.0833 0.26 3.8462

16.00 0.0625 0.29 3.4483

20.00 0.0500 0.41 2.4390

20.00 3.00 0.3333 0.08 12.5000

4.00 0.2500 0.10 10.0000

5.00 0.2000 0.11 9.0909

6.00 0.1667 0.11 9.0909

10.00 0.1000 0.17 5.8824

20.00 0.0500 0.23 4.3478

By-product

(g/L) Glucose (g/L) 1/[S]

Specific growth

rate µ (hr-1) 1/µ

25.00 3.00 0.3333 0.07 14.2857

4.00 0.2500 0.07 14.2857

5.00 0.2000 0.09 11.1111

8.00 0.1667 0.10 10.0000

12.00 0.1000 0.15 6.6667

20.00 0.0500 0.18 5.5556

32.00 3.00 0.3333 0.05 20.0000

4.00 0.2500 0.05 20.0000

5.00 0.2000 0.07 14.2857

8.00 0.1667 0.07 14.2857

12.00 0.1000 0.09 11.1111

20.00 0.0500 0.11 9.0909

Equation

y = 13.556x + 2.3682

y = 28.183x + 3.3185

y = 32.705x + 4.6395

y = 39.386x + 7.9577

-5.0000

0.0000

5.0000

10.0000

15.0000

20.0000

25.0000

-0.2000 -0.1000 0.0000 0.1000 0.2000 0.3000 0.4000

1/[

µ]

1/[s]

Graph of 1/[µ] versus 1/[s]

[I]=0.00 g/L

[I]=20.00 g/L

[I]=25.00 g/L

[I]=32.00 g/L

Linear ([I]=0.00 g/L)

Linear ([I]=20.00 g/L)

Linear ([I]=25.00 g/L)

Linear ([I]=32.00 g/L)

Tabulated value of µmax and ks

Concentration g/L

Y intercept = 1/µmax

µmax, hr-1 X intercept = -1/Ks

Ks

0.00 2.3682 0.4223 -0.1747 5.72

20.00 3.3185 0.3013 -0.1177 8.49

25.00 4.6395 0.2155 -0.1416 7.05

32.00 7.0887 0.1411 -0.1686 5.93

Solution 3b

• LWB plot and calculations in part A suggests that the inhibition type could be noncompetitive because value of µmax decreased proportionally with inhibitor concentration (varying µmax).

• KS values from 5.72 to 8.49 considered small differences

• Slope increment with increased inhibition concentration means that it is a non competitive

• Uncompetitive would have the same slope

• Accuracy of data may not be good. It doesn’t show that it is a 100 % non competitive inhibitor.

• LWB plot, plotting like is just taken best line of graph and not through exact points.

• Experimental data might not be 100% accurate.

• May show uncompetitive type

• Reliability of inhibition constant can be improved through second LWB plot taken the best line of graph.

• Inhibition constant = x-intercept

• Non-competitive type hence Ks should be taken as average to improve Kp value reliability.

Determination of Ks,average

Derivation for non-competitive equation

Kp determination

• µmax,app is the growth rate , µmax with the influence of inhibition which is µmax calculated at part A with different concentration of inhibitor.

• Graph is plotted of each inverse apparent specific growth rate against concentration of inhibitor which is secondary LWB plot to find value of inhibition constant, Kp.

[I], g/L µmax,app (hr-1) 1/µmax,app

or 1/[V'max]

0.00 0.4223 2.368

20.00 0.3013 3.319

25.00 0.2155 4.640

32.00 0.1411 7.087

y = 0.1305x + 1.8419

-1

0

1

2

3

4

5

6

7

8

-15 -10 -5 0 5 10 15 20 25 30 35

1/[

V' m

ax]

[I], g/L

Graph of 1/[V'max] vs [I]

Kp determination

THANK YOU