Biochemical Miscellaneous
- 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.
- 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 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
