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
- 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.
- A callus of 5 g dry weight was inoculated onsemi-solid medium for growth. The dry weight of the callus was found to increase by 1.5 fold after 10 days of inoculation. The growth index of the culture is ______
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Given, Initial dry weight = 5 g
After inoculation for 10 days, dry weight = 1.5 × 5 = 7.5 g
Therefore Growth index of the callus = (Difference or change in weight) / (Originalweight) = (7.5 – 5) / 5 = 0.5Correct Option: C
Given, Initial dry weight = 5 g
After inoculation for 10 days, dry weight = 1.5 × 5 = 7.5 g
Therefore Growth index of the callus = (Difference or change in weight) / (Originalweight) = (7.5 – 5) / 5 = 0.5
- A chemostat is operated at a dilution rate of 0.6 h–1. At steady state, the biomass concentration in the exit stream was found to be 30 g l–1. The biomass productivity (g l–1h–1) after 3h of steady state operation will be ______
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Given, Biomass concentration in the exit stream = 30 g/L
Dilution rate of the chemostat = 0.6 h-1
Therefore, Biomass productivity of the chemostat = 30 × 0.6 gL–1h–1 = 18 gL-1h-1Correct Option: C
Given, Biomass concentration in the exit stream = 30 g/L
Dilution rate of the chemostat = 0.6 h-1
Therefore, Biomass productivity of the chemostat = 30 × 0.6 gL–1h–1 = 18 gL-1h-1
- A bacterial culture (200 µl containing 1.8 × 109 cells) was treated with an antibiotic Z (50 µg per ml) for 4 h at 37°C. After this treatment, the culture was divided into two equal aliquots.
Set A : 100 µl was plated on Luria agar.
Set B : 100 µl was centrifuged, the cell pellet washed and plated on Luria agar.
After incubating these two plates for 24 h at 37°C, Set A plate showed no colonies, whereas the Set B plate showed 0.9 × 109 cells. This experiment showed that the antibiotic Z is
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Bacteriostatic antibiotics limit the growth of bacteria by interfering with bacterial protein production, DNA replication, or other aspects of bacterial cellular metabolism. Since after incubation, plate A showed no colonies and plate B showed 0.9 * 109 colonies (half of the bacterial culture before inoculating), thus it shows that antibiotic Z is bacteriostatic which has stopped the bacteria from reproducing.
Correct Option: A
Bacteriostatic antibiotics limit the growth of bacteria by interfering with bacterial protein production, DNA replication, or other aspects of bacterial cellular metabolism. Since after incubation, plate A showed no colonies and plate B showed 0.9 * 109 colonies (half of the bacterial culture before inoculating), thus it shows that antibiotic Z is bacteriostatic which has stopped the bacteria from reproducing.
