TLS Online TPP Program

#Question id: 14200


A fluidized-bed, immobilized-cell bioreactor is used for the conversion of glucose to ethanol by Z. mobilis cells immobilized in k-carrageenan gel beads. The dimensions of the bed are 10 cm (diameter) by 200 cm. Since the reactor is fed from the bottom of the column and because of CO2 gas evolution, substrate and cell concentrations decrease with the height of the column. The average cell concentration at the bottom of the column is X0 = 45 g/l, and the average cell concentration decreases with the column height according to the following equation:

                             

where Z is the column height (cm). The specific rate of substrate consumption is qS = 2 g S/g cells . h. The feed flow rate and glucose concentration in the feed are 5 l/h and 160 g glucose/l, respectively.

Determine the ethanol concentration in the effluent (g/l . h) if YP/S = 0.48 g ethanol/g glucose

#Section 5: Bioprocess Engineering and Process Biotechnology
  1. 67.9g/l 
  2. 87.5g/l   
  3. 46.2g/l   
  4. 41.9g/l
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TLS Online TPP Program

#Question id: 18955

#Section 7: Recombinant DNA technology and Other Tools in Biotechnology

(i) (RFLPs), (ii) (RAPDs), (iii) (VNTR) DNA, and (iv) (SSRs) are the examples of

TLS Online TPP Program

#Question id: 18965

#Section 7: Recombinant DNA technology and Other Tools in Biotechnology

Select the best algorithm to do pairwise alignment when two proteins are very different in length.

TLS Online TPP Program

#Question id: 18966

#Section 7: Recombinant DNA technology and Other Tools in Biotechnology

Select the best algorithm to do pairwise alignment when two proteins are closely related and very similar in length.

TLS Online TPP Program

#Question id: 18967

#Section 4: Fundamentals of Biological Engineering

If the Reactant iron is combined with Reactant Sulfur to form the product Iron sulfide, then what will be the atomic mass of the product?

TLS Online TPP Program

#Question id: 18968

#Section 4: Fundamentals of Biological Engineering

A continuous process is set up for treatment of wastewater. Each day, 103 kg cellulose and 105 kg bacteria enter in the feed stream, while 102 kg cellulose and 1.5 x 102 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 8 x 102 kg d-1. The rate of bacterial growth is 4x 102 kg d-l; the rate of cell death by lysis is 6 x 102 kg d-1. Write balances for cellulose and bacteria in the system.

TLS Online TPP Program

#Question id: 19021

#Section 4: Fundamentals of Biological Engineering

2SO2+O2 → 2SO3. What is the stoichiometric ratio of SO2 to SO3?