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2003 Course Bioprocess Eng. (Elective I)

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Total No. of Questions : 12] [Total No. of Pages : 3 [3864] - 326 P 1090 B.E. (Chemical Engg.) BIOPROCESS ENGINEERING (2003 Course) (409341) (Elective - I) Time : 3 Hours] [Max. Marks : 100 Instructions to the candidates: 1) Answers to the two sections should be written in separate answer books. 2) Assume suitable data, if necessary. 3) Neat diagrams must be drawn wherever necessary. SECTION - I Q1) Write short notes on the following : a) Protist kingdom. b) Ammylase. c) Co-enzyme. d) Specific growth rate of bacteria. [16] OR Q2) Write short notes on the following : a) Structure of steroids. b) Role of DNA in cell life cycle. c) Limiting nutrient. d) Apo-enzyme. [16] Q3) Explain the manufacturing process for a) Vitamin A and b) Lactic Acid. [16] OR Q4) Explain the manufacturing process for a) Penicillin and b) Vinegar. [16] Q5) Derive the kinetic expression for the following : Km (1) E + S ES; ES k E + P; E + P EP; (2) Kp (3) P.T.O. Where Km and Kp are the thermodynamic dissociation constants for reversible reactions 1 and 3 respectively. k is the kinetic constant for reaction 2. What type of kinetics is represented by the above equations? [18] OR Q6) Data for the enzyme catalyzed reaction S P is as follows : [S] (M) 6.25 x 10 6 7.50 x 10 5 1.00 x 10 4 1.00 x 10 3 1.00 x 10 2 v (nmoles.lit 1. 15.00 56.25 60.00 74.90 75.00 1 min ) a) Estimate Vmax and Km. b) What would v be at [S] = 2.5 x 10 5 M and at [S] = 5.0 x 10 5 M? c) What would v be at 5.0 x 10 5 M if the enzyme concentration were doubled? d) How will you verify that v represents a true initial velocity? [18] SECTION - II Q7) Derive mathematical expressions with the help of Michaelis-Menten inhibition enzymatic kinetics for : a) Noncompetitive inhibition b) Competitive inhibition. [16] OR Q8) a) b) Explain how balanced growth of microbes is needed to be maintained [6] for chemostat and prove that for sterile feed, D = . Operation of a typical CSTR follows the Monod kinetics where max = 0.5h 1 and Ks = 2 g/l. i) At steady state with no cell death, if S0 = 50 g/l and Y = 1 (g cells / g substrates), what dilution rate D will give the maximum total rate of cell production? ii) For the same value of D using tanks of the same size in series, how many vessels will be required to reduce the substrate concentration to 1 g/l? [10] [3864] - 326 -2- Q9) a) b) A marine microorganism contains an enzyme that hydrolyzes glucose6-sulphate (S). The assay is based on the rate of glucose formation. The enzyme in a cell-free extract has kinetic constants of Km = 6.7 x 10 4 M and Vmax = 300 nmoles.lit 1.min 1. Galactose-6-sulphate is a competitive inhibitor (I). At 10 5 M galactose-6-sulphate and 2x10 5 M glucose-6sulphate, v was 1.5 nmoles. lit 1.min 1. Calculate Ki for galactose-6sulphate. [12] Calculate the peak oxygen consumption of specific yeast population in g/(lit.h). Actively respiring yeast population requires 0.32 g oxygen / (hr.g of dry cell mass). Cell population density is 109 cells per ml and single cell volume is 10 10 ml. 80% of active cell mass is water. [4] OR Q10)The steady state substrate and biomass concentrations for a continuous stirred tank fermenter operated at various dilution rates are given below. Given that the fresh feed concentration is 700 mg/l, calculate the values of the Monod constants m and Ks, the yield coefficient Y and the endogenous respiration [16] coefficient Kd. Dilution rate (hr 1) 0.3 0.25 0.2 0.12 0.08 Substrate concentration (mg/l) 45 41 16 8 3.8 Biomass concentration (mg/l) 326 328 340 342 344 Q11)Explain in brief the following : a) Reactor dynamics. b) Immobilization of enzymes. c) Ion exchange chromatography. [18] OR Q12)Explain in brief the following : a) Bubble column bioreactor. b) Fluidized bed bioreactor. c) Continuous sterilization of bioreactor. [3864] - 326 -3- [18]

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