TLS Online TPP Program

#Question id: 14310


Glucose is converted to ethanol by immobilized yeast cells entrapped in gel beads. The specific rate of ethanol production is: qP = 0.2 g ethanol/g-cell-h. The effectiveness factor for an average bead is 0.8. Each bead contains 50 g/L of cells. The voids volume in the column is 40%. Assume growth is negligible (all glucose is converted into ethanol). The feed flow rate is F = 400 l/h and glucose concentration in the feed is S0i = 150 g glucose/l. The diameter of the column is 1 m and the yield coefficient is about 0.49 g ethanol/g glucose. The column height is 4 m. What is the ethanol concentration in the exit stream?

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#Section 5: Bioprocess Engineering and Process Biotechnology
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TLS Online TPP Program

#Question id: 14158

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

BLAST uses a _______ to find matching words, whereas FASTA identifies identical matching words using the _____

TLS Online TPP Program

#Question id: 14159

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

Which of the following statement is not a benefit of FASTA over BLAST?

TLS Online TPP Program

#Question id: 14160

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

In FASTA, For a Z-score > 15, the match can be considered extremely ______ with _____ of a homologous relationship.

TLS Online TPP Program

#Question id: 14161

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

Which of the following is not a description of dynamic programming algorithm?

TLS Online TPP Program

#Question id: 14162

#Section 4: Fundamentals of Biological Engineering

A continuous process is set up for treatment of wastewater. Each day, 105 kg cellulose and 103 kg bacteria enter in the feed stream, while 104 kg cellulose and 1.5 x 104 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 7 x 104 kg d-1. The rate of bacterial growth is 2 x 104 kg d-1 ; the rate of cell death by lysis is 5 x 102 kg d-1. Write balances for cellulose and bacteria in the system.

TLS Online TPP Program

#Question id: 14162

#Section 5: Bioprocess Engineering and Process Biotechnology

A continuous process is set up for treatment of wastewater. Each day, 105 kg cellulose and 103 kg bacteria enter in the feed stream, while 104 kg cellulose and 1.5 x 104 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 7 x 104 kg d-1. The rate of bacterial growth is 2 x 104 kg d-1 ; the rate of cell death by lysis is 5 x 102 kg d-1. Write balances for cellulose and bacteria in the system.