TLS Online TPP Program

#Question id: 14167


Corn-steep liquor contains 2.5 % invert sugars and 50% water; the rest can be considered solids. Beet molasses containing 50% sucrose, 1% invert sugars, 18% water and the remainder solids, is mixed with corn-steep liquor in a mixing tank. Water is added to produce a diluted sugar mixture containing 2% (w/w) invert sugars. 125 kg corn-steep liquor and 45 kg molasses are fed into the tank.

 How much water is required?


#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.