Biochemical Miscellaneous
- 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
- 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 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). ______
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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 = 0.86Correct 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
- Calculate the percentage sequence identity for the pairwise alignment given below._____
H E L L O –
Y E L L O W
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Given, two sequences HELLO- and YELLOW
The % identity between the two sequences would be : No. of identical positions / Total no. of positions
No. of identical positions = 4 (ELLO); Total no. of positions = 6
Therefore, percentage identity = 4 / 6 × 100 = 66.667 %Correct Option: B
Given, two sequences HELLO- and YELLOW
The % identity between the two sequences would be : No. of identical positions / Total no. of positions
No. of identical positions = 4 (ELLO); Total no. of positions = 6
Therefore, percentage identity = 4 / 6 × 100 = 66.667 %
- A batch bioreactor is to be scaled up from 10 to 10,000 liters. The diameter of the large bioreactor is 10 times that of the small bioreactor. The agitator speed in the small bioreactor is 450 rpm. Determine the agitator speed (rpm) of the large bioreactor with same impeller tip speed as that of the small bioreactor. ______
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Let the agitator speed in the large bioreactorbe VL
Now, we have for scale up : VL × DL = VS × DS
Where, DL = diameter of the large bioreactor,
VS = agitator speed of the small bioreactor = 450 rpm,
DS = diameter of the small bioreactor,
n = scale up ratio = 10,000 / 10 = 1000
Now, DL = 10 DS
Therefore, VL × 10 DS = 450 × DS
⇒ VL = 45
Hence the agitator speed of the large bioreactor is 45 rpm.Correct Option: A
Let the agitator speed in the large bioreactorbe VL
Now, we have for scale up : VL × DL = VS × DS
Where, DL = diameter of the large bioreactor,
VS = agitator speed of the small bioreactor = 450 rpm,
DS = diameter of the small bioreactor,
n = scale up ratio = 10,000 / 10 = 1000
Now, DL = 10 DS
Therefore, VL × 10 DS = 450 × DS
⇒ VL = 45
Hence the agitator speed of the large bioreactor is 45 rpm.
