Biochemical Miscellaneous

Biochemical Miscellaneous

Easy Questions

Biochemical

Biochemical Miscellaneous

Direction: During sterilization of a fermentation medium in a given bioreactor ∇heating = 12.56, ∇cooling = 7.48 and the ​total value of ∇ required for whole sterilization process is 52, where ∇ is the design criteria. ​

  1. What is the holding period (min) at a k value of 3.36 min–1? ​








  1. View Hint View Answer Discuss in Forum

    Given k value = 3.36 min–1 ​
    Holding value of design criteria, = 31.96
    ​Therefore, time of holding, ​  t = ∇holding / k ​​​
    = 31.96 / 3.36 min = 9.512 min.

    Correct Option: B

    Given k value = 3.36 min–1 ​
    Holding value of design criteria, = 31.96
    ​Therefore, time of holding, ​  t = ∇holding / k ​​​
    = 31.96 / 3.36 min = 9.512 min.

  1. What is the value of ∇holding? ​​








  1. View Hint View Answer Discuss in Forum

    Given,  ∇heating = 12.56,  ∇cooling = 7.48 ​
    Now we have,  ∇total = ∇heating + ∇cooling + ∇holding
    ⇒ ​​​52 = 12.56 + 7.48 +​ ∇holding
    ⇒  ​​  ∇holding = 31.96  

    Correct Option: A

    Given,  ∇heating = 12.56,  ∇cooling = 7.48 ​
    Now we have,  ∇total = ∇heating + ∇cooling + ∇holding
    ⇒ ​​​52 = 12.56 + 7.48 +​ ∇holding
    ⇒  ​​  ∇holding = 31.96  

  1. A fed batch culture was operated with intermittent addition of glucose solution at a flow rate of 200 ml h–1. The values of Ks, µm and D are 0.3 g l–1, 0.4 h–1 and 0.1 h–1, respectively. Determine the concentration of growth limiting substrate (gl–1) in the reactor at quasi-steady state. _______








  1. View Hint View Answer Discuss in Forum

    Given, flow rate for the glucose solution, F = 200 mL/hr
    ​​KS = 0.3 g/L; µm = 0.4 h-1; D = 0.1 h-1​
    Therefore, Volume of the reactor, ​​V = F / D = 200 / 0.1 mL = 2000 mL = 2L ​
    We have, from Monod’s equation of substrate consumption µ = (µmax * S)/ (KS + S)
    where, ​​µ = rate of substrate consumed per unit time.  
    µmax = maximum rate of substrate consumed.  
    KS = Monod’s constant.
    S = steady state substrate concentration.
    In steady state, mass balance equation is given by :-
    Input – Output – Consumption = 0 ​⇒​(F / V)*SR – µ = 0 ​
    ⇒​D(S) – {(µmax*S)/(KS + S)} = 0 ​
    Therefore, ​D(S) = {(µmax*S)/(KS + S)} = {(µmax*S)/(KS + S)}
    = {(0.4S) / (0.3 + S)} ​
    ⇒​ DS = 0.4S / (0.3 + S) ​
    ⇒ ​0.1 = 0.4S / (0.3 + S) ​
    ⇒​ 0.4S = 0.03 + 0.1S ​
    ⇒​ 0.3S = 0.03 ​
    ⇒​ S = 0.1 ​
    Therefore, the concentration of the limiting substrate = 0.1 g/L

    Correct Option: A

    Given, flow rate for the glucose solution, F = 200 mL/hr
    ​​KS = 0.3 g/L; µm = 0.4 h-1; D = 0.1 h-1​
    Therefore, Volume of the reactor, ​​V = F / D = 200 / 0.1 mL = 2000 mL = 2L ​
    We have, from Monod’s equation of substrate consumption µ = (µmax * S)/ (KS + S)
    where, ​​µ = rate of substrate consumed per unit time.  
    µmax = maximum rate of substrate consumed.  
    KS = Monod’s constant.
    S = steady state substrate concentration.
    In steady state, mass balance equation is given by :-
    Input – Output – Consumption = 0 ​⇒​(F / V)*SR – µ = 0 ​
    ⇒​D(S) – {(µmax*S)/(KS + S)} = 0 ​
    Therefore, ​D(S) = {(µmax*S)/(KS + S)} = {(µmax*S)/(KS + S)}
    = {(0.4S) / (0.3 + S)} ​
    ⇒​ DS = 0.4S / (0.3 + S) ​
    ⇒ ​0.1 = 0.4S / (0.3 + S) ​
    ⇒​ 0.4S = 0.03 + 0.1S ​
    ⇒​ 0.3S = 0.03 ​
    ⇒​ S = 0.1 ​
    Therefore, the concentration of the limiting substrate = 0.1 g/L

  1. The maximum cell concentration (g l–1) expected in a bioreactor with initial cell concentration of 1.75 g l–1 and an initial glucose concentration of 125 g l–1 is (Yx/s = 0.6 g cell/g substrate)_______ ​








  1. View Hint View Answer Discuss in Forum

    Given, Initial cell concentration, X = 1.75 g/L ​
    Initial concentration of substrate glucose, ​​S = 125 g/L
    ​Given, Yield of biomass, YX/S = 0.6 g/g
    Now, Concentration of biomass produced ​​​= YX/S × S = 125 × 0.6 g/L = 75 g/L ​
    Therefore, maximum concentration of cell expected in the bioreactor ​​​= 75 + 1.75 g/L = 76.75 g/L

    Correct Option: B

    Given, Initial cell concentration, X = 1.75 g/L ​
    Initial concentration of substrate glucose, ​​S = 125 g/L
    ​Given, Yield of biomass, YX/S = 0.6 g/g
    Now, Concentration of biomass produced ​​​= YX/S × S = 125 × 0.6 g/L = 75 g/L ​
    Therefore, maximum concentration of cell expected in the bioreactor ​​​= 75 + 1.75 g/L = 76.75 g/L

  1. In a batch culture, the specific rate of substrate utilization is 0.25 g (g cell mass)–1 h–1 and specific rate of product formation is 0.215 g (g cell mass)–1 h–1. Calculate the yield of product from the substrate (Yp/s). ______ ​








  1. View Hint View Answer Discuss in Forum

    Given, for a batch culture ​
    Specific rate of substrate utilization = 0.25 g/g/h ​
    Specific rate of product formation = 0.215 g/g/h ​
    Yield of product (YP/S) = 0.215/.25 = 0.86

    Correct Option: A

    Given, for a batch culture ​
    Specific rate of substrate utilization = 0.25 g/g/h ​
    Specific rate of product formation = 0.215 g/g/h ​
    Yield of product (YP/S) = 0.215/.25 = 0.86