Ministry of Science and Technology

Ministry of Science and Technology

Department of Technical and Vocational Education

B.E (Chemical) 2006

Final Examination

23-10-2006

8:30am- 11:30am

ChE 05022 Biochemical Engineering Answer any five questions1. Derive the rate equation for the following enzyme reaction using the Briggs-Haldane approach.k1

k3S+E

ES

ES

P+|Ek22. A substance is converted to a product by the catalytic action of enzyme. Assume that the Michaelis-Menten kinetic parameters for this enzyme reaction are:

KM = 0.03 mol/l, rmax =13 mol/l-min

(a) What should be the size of a steady-state CSTR to remove 95 percent of incoming substrate ( Cso = 10 mol/l ) with a flow rate of 10 l/hr?

(b) What should be the size of the reactor if you employ a plug-flow reactor instead of the CSTR in part (a)?

3. Roughly sketch a prokaryotic cells, label its parts, and state a function for each of these.

4. What are the basic procedures of recombinant DNA techniques? Explain briefly these techniques with the help of flow chart.

5. Explain in detail the growth cycle of unicellular micro-organism for batch cultivation.

6. Aiba et al.(1968) reported the results of a chemostat study on the growth of a species strain of bakers yeast as shown in the following table. The inlet stream of the chemostat did not contain any cells or products.

Dilution Inlet Steady-state Steady-state Steady-state

RateGlucose Conc. Glucose Conc. Ethanol Conc. Cell Conc. D,hr-1 Csi ,g/l Cs,g/l Cp ,g/l Cx ,g/l 0.084 21.5 0.054 7.97

2.00

0.100 10.9 0.079 4.70

1.20

0.160 21.2 0.138 8.57

2.40

0.198 20.7 0.186 8.44

2.33

0.242 10.8 0.226 4.51

1.25

(a) Find the rate equation for cell growth.

(b) Find the rate equation for product (ethanol) formation.

3. Prokaryotic cells

Typical prokaryotic cells

Typical prokaryotic cells

The prokaryotic cell is unit of structure, two group microbial, bacteria and blue green algae. Prokaryotic cell is simple and small. Prokaryotic cell is not compartmentalized by unit membrane. Prokaryotic cell have two structurally distinguished two internal region: cytoplasm and nuclear region. Cytoplasm contains graing dark spot due to the present of ribosome.

Ribosome is the site of important biochemical reaction for protein synthesis. The nucleur contain deoxyribose nucleuic acid (DNA) ,which contain genetic information that determine the production of protein and structure .

The prokaryotic cell is surrounded by cell wall and a cell membrane. The cell wall is considerable thicker than cell membrane, and it protect cells from external influence. The cell membrane serve as the surface which other cells components attack and important cell function take place. 4. The basic procedures of recombinant DNA techniques have two factor. They are

(1) to join the DNA segment to the molecule and they can be replicated . (2) to provide environment that allow the reproduction of join DNA molecules . The produce for the flow chart of recombinant DNA techniques are

Fig : The procedure of recombinant DNA .

Figure show the procedure of recombinant DNA techniques. The plasmid cut in definite size with restriction enzyme. The DNA of the foreign cleaved with restriction enzyme. Some of fragment have interest gene. The plasmid and genome fragment mixed and joined by the DNA Ligases. The recombinant plasmids are introduced into the bacteria and coculativation of plasmid and bacteria.s1 .solution

k1

k3S+E

ES

ES

P+|E

k2The intermediate of reaction with respect to time is neglected.

dCES= 0

dt

rCES = k1CsCE- k2 CES- k3 CES = 0

(1)

k1CsCE = k3 CES + k2CES k1CsCE = (k2 +k3 ) CES CE = (k2 + k3 ) CES

k1Cs

rp = k3 CES (2)By enzyme balance,

CEO = CE + CES

= (k2 + k3 ) CES

k1Cs = ( k2 + k3 +1) CES

k1Cs

CEO = (k2 + k3 + k1Cs) CES

k1Cs

By dividing by k1

CES = CEOCS

k2 + k3 +Cs

k1

By substituting of eqn: (2)

rp = k3 CES

= k3 CEO CS

k2 + k3 +Cs

k1

Comparing with rp = rmax Cs

kM +CS

k3 CEO CS = rmax CS

k2 + k3 +Cs kM +CS

k1

( k2+k3 ( k2

k1

k1

rmax = k3CEO kM = k2+k3 ( k2

k1 k1

2. Solution

kM = 0.03 mol/Lrmax =13mol ( 60 min = 780 mol L min1 hr L hr(a) v =?

xs =95%

Cso =10 mol / L

F = 10 L / hr

CS= CSO (1_ xs)

= 10 (1 _ 0.95) = 0.5 mol /L

At the CSTR,

Input _ Output + general = 0FCSO _ FCS +rsv = 0

F (CSO _CS) = _ rsv

F (CSO _Cs) = rpv F (CSO _CS) = rmax Cs V

KM +CS

F = rmax Cs

V (KM+CS)(CSO_CS)

F = 780* 0.5

V (0.03+0.5) (10-0.5)

F= 390

V 0.53 * 9.5

F = 77.46 hr-1

V

V = F

77.46hr-1 V = 10 L/hr

77.46

V = 0.129 liter.

2 .(b)

The plug flow reactor , V = ? _dCs = rmax Cs

dt KM +CS Cs

t_ ( dcs = rmax Cs = Cs

( rmax

Cso dtKM +CS KM +CS 0

Cso

t( KM +CS dCs = ( rmax tCs Cs 0Cso Cso( kM dCs + ( dCs = rmax t Cs Cs

kMlnCso + ( Cso _Cs ) = rmaxt ( plus = t batch )

kMlnCso + ( Cso _Cs ) = rmax

0.03 ln 10 + (10-0.5) = 780

0.5

0.0899+ 9.5= 780

9.95899 = 780

= 0.0123hr

= V / F = 0.0123hr V = 0.0123

10 L/hr hr

V = 0.123 Liter

6. Soln:

(a) rx = ?

(b) rp = ?

=

EMBED Equation.3

=

EMBED Equation.3 +

y = m x + c D = hr-1 CS g/l

0.084 0.054 11.9 18.5

0.1 0.16 10 12.6

0.16 0.13 6.25 7.25

0.198 0186 5.05 5.38

0.242 0.226 4.13 4.42

From graph,

C = = 1.6

max = = 0.625

m = = = 0.667

ks = 0.42 g/l

6. (a) rx = ?

rx =

rx =

6. (b) rp = ?

rp =

EMBED Equation.3

=

EMBED Equation.3 = =

=

=

= 0.27rp =

rp =

1.(a) The rates of product formation and substance of consumption are

= k3CES

-= k1CsCE k2CES ( (1)Assume that the change of CES with time, d CES/dt, is neglible compared to that of CP or CS .

= k1CSCE k2 CES k3 CES ( 0 ( (2)Substituting Eq (1) into Eq (2) confirms that the rate of product, formation and that of the substrate consumption are the same, that is ,

r = = -= k3 CES

If we assume that the total enzyme contents are conserved,

CEO = CE + CES ( (3) Substituting Eq (3) into Eq (2) for CE, and rearranging for CES

CES =

r = = -= =

While in the Briggs- Haldane approach, it is equal to ( k2 + k3 ) / k1 .

2. KM = 0.03 mol/lit Umax = 13 mol/lit

CSO = 10 mol/lit

F = 10 L / hr ( =Lit / min

(a) V =? , CS = 0.05 CSO = 0.5 mol / lit

(a) CSTR

Input output + generation = accumulation

FCSO _ FCS + (S V = V = 0

F (CSO_ CS) = _(S V

=

=

V =

=

= 0.129 Lit.(b) V =?

CS For plug flow reaction

V =0.123lit

3. The prokaryotic cell is the unit of structure in two microbial groups: bacteria and blue-green algae. The prokaryotic is small and simple , as shown in fig , which is not compartmentalized by unit membrane systems .The cell has only two structurally distinguishable internal regions: cytoplasm and nuclear region (or nucleoplasm ).The cytoplasm has grainy dark spots as a result of its content of ribosomes, which are composed of protein and ribonucleic acid (RNA) . The ribosome is the site of important biochemical reactions for protein synthesis. The nuclear region is of irregular shape , sharply segregated even though it is not bounded by membrane .The nuclear region contains deoxyribonucleic acid (DNA) ,which contains genetic information that determines the production of proteins and other cellular substances and structures .

The prokaryotic cell is surrounded with a cell wall and a cell membrane.

The cell wall, considerably thicker than the cell membrane, protects the cell from external influences. The cell membrane (or cytoplasmic membrane) is a

selective barrier between the interior of the cell and the external environment .

The largest molecules known to cross this membrane are DNA fragments and low-molecular-weight proteins. The cell membrane can be folded and extended into the cytoplasm or internal membranes. The cell membrane serves as the surface onto which other cell substances attach and upon which many important cell functions take place.

4

Fig, Method for the production of recombinant DNA

Fig shows the overall procedure for the production of recombinant DNA .The plasmid is cut at a number of defined sites with a restriction enzyme. The DNA of a foreign genome is cleaved with a restriction enzyme. Some of the resulting fragments may have the gene of interest. Plasmids and genome fragments are mixed and joined by DNA ligase. The recombinant plasmids are introduced into bacteria by cocultivation of plasmids and bacteria.

5. Growth Cycle for Batch Cultivation

If you inoculate unicellular microorganisms into a fresh sterilized medium and measure the cell number density with respect to time and plot it, you may find that there are six phases of growth and death,as shown in Fig. They are:

1. Lag phase: A period of time when the change of cell number is zero.

2. Accelerated growth phase: The cell number starts to increase and the division rate increases to reach a maximum.

3. Exponential growth phase: The cell number increases exponentially as the cells start to divide. The growth rate is increasing during this phase, but the division rate which is proportional to d ln Cno / dt, is constant at its maximum value, as illustrated in fig .

4. Decelerated Growth phas: After the growth rate reaches a maximum, it is followed by the deceleration of both growth rate and the division rate .

5. Stationary phase: The cell population will reach a maximum value and will not increase any further.

6. Death phase: After nutrients available for the cells are depleted, cells will start to die and the number of viable cells will decrease.

1 Lag Phase

The lag phase (or initial stationary, or latent) is an initial period of cultivation during which the change of cell number is zero or negligible. Even though the cell number does not increase, the cells may grow in size during this period.

The length of this lag period depends on many factors such as the type and age of the microorganisms, the size of the inoculum, and culture conditions. The lag usually occurs because the cells must adjust to the new medium before growth can begin. If microorganisms are inoculated from a medium with a low nutrient concentration to a medium with a high concentration, the length of the lag period is usually long. This is because the cells must produce the enzymes necessary for the metabolization of the available nutrients. If they are moved from a high to a low nutrient concentration, there is usually no lag phase.

Another important factor affection the length of the lag phase is the size of the inoculum. If a small amount of cells are inoculated into a large volume, they will have a long lag phase. For large-scale operation of the cell culture, it is our objective to make this lag phase as short as possible. Therefore, to inoculate a large fermenter, we need to have a series of progressively larger seed tanks to minimize the effect of the lag phase.

Fig Typical growth curve of unicellular organisms: (A) lag phase; (B) accelerated growth phase; (C) exponential phase; (D) decelerated growth phase; (E) stationary phase;(F) death phase.

At the end of the lag phase, when growth begins, the division rate increases gradually and reaches a maximum value in the exponential growth period, as shown by the rising inflection at B in fig. This transitional period is commonly called the accelerated growth phase and is often included as a part of the lat phase.

In unicellular organisms, the progressive doubling of cell number results in a continually increasing rate of grown in the population. A bacterial culture undergoing balanced growth mimics a first-order autocatalytic chemical reaction. Therefore, the rate of the number density (Cn) of bacteria present at that time:

where the constant is known as the specific growth rate.The specific growth rate should not be confused with the growth rate, which has different units and meaning. The growth rate is the change of the cell number density with time, while the specific growth rate is

The growth of microbial populations is normally limited either by the exhaustion of available nutrients or by the accumulation of toxic products of metabolism. As a consequence, the rate of growth declines and growth eventually stops. At this point a culture is said to be in the stationary phase. The transition between the exponential phase and the stationary phase involves a period of unbalanced growth during which the various cellular components are synthesized at unequal rates. Consequently, cells in the stationary phase have a chemical composition different from that of cells in the exponential phase.

The stationary phase is usually followed by a death phase in which the organisms in the population die. Death occurs either because of the depletion of the cellular reserves of energy, or the accumulation of toxic products. Like growth, death is an exponential function. In some cases, the organisms not only die but also disintegrate, a process called lysis.

6. (a) Cxi= 0 , Csi = 0

Plot

Intercept

D = hr-1 CS g/l

0.084 0.054 11.9 18.5

0.1 0.16 10 12.6

0.16 0.13 6.25 7.25

0.198 0186 5.05 5.38

0.242 0.226 4.13 4.42

Slope = , Intercept = 1.2

6. (b) Material balance for Pdt in CSTF ,

EMBED Equation.3

EMBED Equation.3

For S.S CSTF with sterile feed,

,

From Manod,

=

straight line

. . .

. .. .....

. .. .. ..

. ..

Cell wall

Cell membrane

Ribosome

Nuclear region

cytoplasm

DNA Foreign

Stricky end

DNA Ilagse

Tranaformation

Transformation cell

CS0

Cs

Plug Flow

Cs

CS0

. .. ....

. .. .....

. .. .. ..

. ..

Cell wall

Cell membrane

Ribosome

Nuclear region

cytoplasm

Cs

CS0

Transformed cell

DNA Ilagse

'Stricky end'

Foreign DNA

t

Restriction enzyme

Tranaformation

Plasmld

8

10

12

A

B

D

C

E

F

PAGE 6

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