Biochemical Miscellaneous
- Since mammalian cells are sensitive to shear, scale-up of a mammalian cell process must consider, among other parameters, the following (given N = rotations/time, D = diameter of impeller)
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Shear speed of an impeller along with the diameter of the impeller blade have profound impact on the shear stress generated in the bioreactor system. Mammalian cells are very shear sensitive. So, for scale up of a small scale bioreactor where mammalian cells are used as feed, the major parameters to be considered would be the shear speed and agitator diameter. This is because, while scaling up, impeller speed has to be considered for controlling the shear speed which is given by : TI = πND
Correct Option: A
Shear speed of an impeller along with the diameter of the impeller blade have profound impact on the shear stress generated in the bioreactor system. Mammalian cells are very shear sensitive. So, for scale up of a small scale bioreactor where mammalian cells are used as feed, the major parameters to be considered would be the shear speed and agitator diameter. This is because, while scaling up, impeller speed has to be considered for controlling the shear speed which is given by : TI = πND
- A T-flask is seeded with 105 anchorage-dependent cells. The available area of the T-flask is 25 cm2 and the volume of the medium is 25 ml. Assume that the cells are rectangles of size 5 µm × 2 µm. If the cells grow to monolayer confluence after 50 h, the growth rate in number of cells/(cm2.h) is ____ × 105 .
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Given, Seeding of T-flask with = 105 no. of anchorage dependent cells
Area of flask available = 25 cm2 and volume of flask = 25 mL
Size of cells = 5µm × 2µm
Area of the T-flask covered by a single cell = 10 µm2 = 10 × 10-8 cm2
Now, anchorage dependent cells multiply every hour per unit area
So, the growth rate of the cells = 2 × 105 cells/cm2h Correct Option: B
Given, Seeding of T-flask with = 105 no. of anchorage dependent cells
Area of flask available = 25 cm2 and volume of flask = 25 mL
Size of cells = 5µm × 2µm
Area of the T-flask covered by a single cell = 10 µm2 = 10 × 10-8 cm2
Now, anchorage dependent cells multiply every hour per unit area
So, the growth rate of the cells = 2 × 105 cells/cm2h
- Consider a continuous culture provided with a sterile feed containing 10 mM glucose. The steady state cell density and substrate concentration at three different dilution rates are given in the table below.
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At dilution rate = 5h-1, cell density = 0. This means cell wash out occurs at this dilution rate
The mass balance on the microorganisms in a CSTR of constant volume is :
[Rate of accumulation of cells, g/s] = [Rate of cells entering, g/s] – [Rate of cells leaving, g/s] + [Net rate of generation of live cells, g/s]
So, maximum specific growth rate µmax = 0.8 h-1 Correct Option: A
At dilution rate = 5h-1, cell density = 0. This means cell wash out occurs at this dilution rate
The mass balance on the microorganisms in a CSTR of constant volume is :
[Rate of accumulation of cells, g/s] = [Rate of cells entering, g/s] – [Rate of cells leaving, g/s] + [Net rate of generation of live cells, g/s]
So, maximum specific growth rate µmax = 0.8 h-1
- To optimize the bioreactor system, which one of the following conditions is least important for anaerobic fermentation ?
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anerobic fermentation doesn’t need agitation to maintain oxygen supply.
Correct Option: A
anerobic fermentation doesn’t need agitation to maintain oxygen supply.
Direction: A bioreactor of working volume 50 m3 produces a metabolite (X) in batch culture under given operating conditions from a substrate (S). The final concentration of metabolite (X) at the end of each run was 1.1 kg m–3. The bioreactor was operated to complete 70 runs in each year.
- What will be the annual output of the system (production of metabolite (X)) in kg per year ?
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In single run metabolite (X) concentration is =1.1 kg m–3.
Since, total number of runs = 70
Total concentration of (X) = 70 × 1.1 kg m–3 run–1
= 70 × 1.1 kg m–3 = 77 kg m–3
Now, total volume of batch culture = 50 m3.
Total amount will be = 50 m3 × 77 kg m–3
= 3850 kg per year Correct Option: B
In single run metabolite (X) concentration is =1.1 kg m–3.
Since, total number of runs = 70
Total concentration of (X) = 70 × 1.1 kg m–3 run–1
= 70 × 1.1 kg m–3 = 77 kg m–3
Now, total volume of batch culture = 50 m3.
Total amount will be = 50 m3 × 77 kg m–3
= 3850 kg per year
