# Indian Institute of Technology Kharagpur (IIT-K) 2007 B.Tech Biotechnology Biochemical reaction engineering & bioenergetics - Question Paper

INDIAN INSTITUTE OF TECHNOLOGY, KHARAGPUR Time: 3 Hrs.

Date of Examination................... End-Autumn Semester 2007-2008 .FN/AN Full Marks : 50 |
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Subject No. BT21105 Subject: Biochemical Reaction Engineering & Bioenergetics No. of Students : 30 of the Department of Biotechnology

Instruction : Answer all questions from Group-A and any two questions from group-B. Answer to the point. Assume suitable data with necessary justification, if required.

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2. a) In PS II which molecule is electron donor which molecules are electron acceptor. What is function of

Peso molecule on PS II? 1+1 |

The release of one dioxygen molecule requires the transfer of 4 electrons from 2 molecules of water to 2 molecules of NADP^{+}. Given the Emj of 2H2O 2H^{+} + O2 couple as +0.82 V and E_{m},7 of NADP^{+}+ 2e +2H^{+} NADPH couple as -0.32 V. Calculate the net thermodynamic efficiency of the electron transfer

steps of photosynthesis.

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3. i) Enumerate the equations related to enthalpy, entropy and Gibbs free energy with heat capacity and temperature, ii) Define T_{m} of protein unfolding and how will you calculate fraction unfolded from a typical thermograph of protein unfolding due to increase of temperature. Iii) What is the significance of

the measurement of AC_{P} of protein unfolding.

2 + 2+1

4. a)Binding constant ( K) of ligand( X) with macromolecule( M) is accurately determined by titration calorimeter. Enumerate the equation related to heat change ( dQ) with change in [MX} concentration, molar enthalpy and cell volume. 1

b) How the dimensionless parameter C is related with macromolecular concentraion and binding constant (K). What is the range of C by which binding constant ( K) could be determined by using titration calorimeter. 2

c)Describe in brief the thermodynamic signature optimization process for the selection of HIV protease inhibitor drug. 2

Group-B

5. a) An enzyme is immobilized on the surface of a non-porous solid matrix. Assuming that external

mass-transfer resistance for substrate is not negligible and that the Michaelis-Menten equation describes the intrinsic kinetics. Derive an expression which indicates the explicit form of the coefficients in a Lineweaver-Burk plot. Find out the values of v_{ma}x, K_{m} and k_{m}(mass-transfer coefficient)?

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b) Glucose is converted to fructose by using immobilized glucose isomerase. Find out the height of the immobilized enzyme column? Following data are given:

1. Diameter of the column = 5 cm.

2. Particle size 30/40 mesh (about 0.71 mm average diameter, d_{p}),

3. Glucose concentration in feed at 60C = 500 g/1,

4. Glucose conversion efficiency = 60% w/w,

5. Feed viscosity = 3.6 c.p. at 60C,

6. Feed density = 1.23 g/ml at 60C,

7. Substrate diffusivity = 0.21 x 10'^{5} cm^{2}/sec at 60C,

8. Void fraction = 0.35. 7

(a) What do you mean by DamkOhler number and effectiveness factor of an immobilized enzyme process? Find out the mathematical correlation between them. Explain the diffusion-limited regime of the immobilization process. 2+2+1.5

b) Write the Monod chemostat models. Derive the equation for finding out the cell mass and substrate concentration in a chemostat under steady state conditions. Write the mathematical expressions for the D_{max} and D_{waS}h_{0}ut. Explain the importance ofD_{waS}hout for the operation of a CSTR system.

2+1+2+2

y, i) Candida utilis utilizes xylose at any concentration constitutively with a maximum growth rate of

0.31 h'^{1} and has a saturation constant (Ks) of 2.6 g/I and a cell yield on substrate (IVs) of 0.5 g/g. C. utilis can also metabolize galactose (with Ks = 0.8 g/1 and Y_{x/S} = 0.5 g/g), but can only metabolize galactose at xylose concentrations at or below 0.5 g/1. If we have a feed stream that contains 10 g/1 of xylose and 30 g/1 of galactose entering a chemostat containing C. utilis. Find the following:

a. D (h'^{1}) at which galactose begins to be metabolized.

a. Dilution rate for maximum cell productivity (D_{opt})

b. Maximum ceil productivity

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ii) D-(-)-4-hydroxyphenylglycin is the optically active intermediate in the synthesis of the broad-spectrum antibiotics amoxicilline. This intermediate is among others produced from a hydantoin derivative by means of the enzyme hydantoinase. The hydantoin derivative is poorly soluble in water, about I kg m'^{3}. The price of the substrate is cost determining and degree of conversion should therefore be very high, at least 99%.

Calculate the volume needed to produce 100 kg of product per day by immobilized hydantoinase in a

- batch reactor 2.5

- CSTR 2

- Plug-flow reactor. 2

Data:

Michaelis-Menten reaction kinetics with v_{ma}x = 1.5 x 10^{-4} kg s'^{3} m'^{3} biocatalyst and K_{m} = 5 x 10'

^{3} kg m^{3}

Yp/s = 1 kg kg^{1} Degree of conversion 99%

Down-time for batch reactor 12 h

The activity of the biocatalyst can be assume to be constant in time

The reactor contain 0.1 m^{3} immobilized biocatalyst per m^{3}, except the plug-flow reactor, which is packed with 0.5 m^{3} immobilized biocatalyst per m^{3} reactor.

Attachment: |

Earning: Approval pending. |