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.
- What is the holding period (min) at a k value of 3.36 min–1?
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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.
- 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. _______
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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/LCorrect 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
- What is the value of ∇holding?
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Given, ∇heating = 12.56, ∇cooling = 7.48
Now we have, ∇total = ∇heating + ∇cooling + ∇holding
⇒ 52 = 12.56 + 7.48 + ∇holding
⇒ ∇holding = 31.96Correct 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
- 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)_______
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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/LCorrect 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
- In an exponentially growing batch culture of Saccharomyces cerevisiae, the cell density is 20 gl–1 (DCW), the specific growth rate (µ) is 0.4h–1 and substrate uptake rate (ν) is 16 gl–1h–1. The cell yield coefficient Yx/s will be
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Yield coefficient = mass of new cells formed / mass of substrate consumed
New cells mass formed = cell density x specific growth rate = 20 × 0.4 = 8
Substrate consumed = 16 g/l/h Yield coefficient = 8 = 0.5 16
Correct Option: D
Yield coefficient = mass of new cells formed / mass of substrate consumed
New cells mass formed = cell density x specific growth rate = 20 × 0.4 = 8
Substrate consumed = 16 g/l/h Yield coefficient = 8 = 0.5 16
