41. Examine the data given below in the table on purification of a protein X
    
42. 1. 25 and 3

    
    
    

    2. 30 and 20

    
    
    

    3. 20 and 30

    
    
    

    4. 75 and 30

    
    
    

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   NA

   Correct Option: B

   NA

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42. Phytase an enzyme produced by Aspergillus niger can be adsorbed on microcrystalline cellulose powder (MCCP). The adsorption follows a Langmuir isotherm and the maximum concentration of the protein that can be obtained on the adsorbent is 70 mg cm^–3. At a concentration of 50mgl^–1 of protein in the solution the concentration of protein on the adsorbent reaches 35 mg cm^–3. It is desired to recover 90% of the addition of MCCP to the solution. The concentration of the protein adsorbed on the solid at equilibrium will be
    

1. 1. 21.4 mg cm^–3

   
   
   

   2. 214 mg cm^–3

   
   
   

   3. 2.14 mg cm^–3

   
   
   

   4. None of these

   
   
   

   5. NA

   
   
   

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   NA

   Correct Option: E

   NA

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43. Inversion of sucrose by immobilized invertase follows a substrate inhibited kinetics. The reaction rate (v) in mol m^–3 hr^–1 can be expressed
    expressed as V = 800 [S]/ {400 + 50 [S] + [S]²}, where [S] is the sucrose concentration.
    The immobilized invertase preparation is used in a CSTR with 100 mol m^–3 sucrose concentration in the feed stream. If the reaction velocity passes through a maxima at [S] = 20 mol m^–3, the feed flow rate for a reactor volume of 1m³ to get the maximum productivity from the reactor should be
    

1. 1. 0.11 m³ hr^–1

   
   
   

   2. 1.10 m³ hr^–1

   
   
   

   3. 5.05 m³ hr^–1

   
   
   

   4. None of these

   
   
   

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   V =	800|g|
   	400 - 500[s] + [s]²

   
   S0 = 100 mol/m³ → V₀ = 5.195 ml/m³ hr.
   i.e.  V_max when S = 20 ml/m³,
   ∴ V_max = 8.889 ml/m³.hr. V = (dp/dt)D = F/V
   Now, K_m = V_max/2 = 4.444
   Now,
   V₀ = V_max = S/K_m + S = 7.27 ml/m³. hr
   

   Correct Option: D

   V =	800|g|
   	400 - 500[s] + [s]²

   
   S0 = 100 mol/m³ → V₀ = 5.195 ml/m³ hr.
   i.e.  V_max when S = 20 ml/m³,
   ∴ V_max = 8.889 ml/m³.hr. V = (dp/dt)D = F/V
   Now, K_m = V_max/2 = 4.444
   Now,
   V₀ = V_max = S/K_m + S = 7.27 ml/m³. hr
   

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44. An enzyme following Michaelis – Menten kinetics with V_m = 2.5 mmol m^–3 s^–1 and K_m = 5.0 mM was used to carry out the reaction in a batch stirred reactor. Starting with an initial substrate concentration of 0.1 M. The time required for 50% conversion of the substrate will be about
    

1. 1. 01 hr

   
   
   

   2. 06 hr

   
   
   

   3. 02 hr

   
   
   

   4. 12 hr

   
   
   

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   By MM equation V = V_m .S/K_m + S.

   Correct Option: B

   By MM equation V = V_m .S/K_m + S.

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45. The two columns given below indicate some of the fermentation products and the microbial cultures used for their production. Identify the correct set of groups from the four options.
    Fermentation
    Microbial
    products
    cultures
    (A) Ethanol (P) Aspergillus niger
    (B) Streptomycin (Q) Zymomonas mobiles
    (C) Citric acid (R) Streptomyces griseus
    (D) Cellulase (S) Trichoderma reesei
    

1. 1. A-P; B-Q; C-R; D-S

   
   
   

   2. A-Q; B-R; C-P; D-S

   
   
   

   3. A-Q; B-R; C-S; D-P
      

   
   
   

   4. A-S; B-R; C-Q; D-P

   
   
   

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   A. niger can ferment Citric Acid, Ethanol can be produced by Z. mobilis, Streptomycin is produced by S.griseus, Cellulose is produced by T. ressei.
   

   Correct Option: B

   A. niger can ferment Citric Acid, Ethanol can be produced by Z. mobilis, Streptomycin is produced by S.griseus, Cellulose is produced by T. ressei.
   

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