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TLS Online TPP Program

#Question id: 10142

Which of the following statements is INCORRECT?

#Section 2: General Biology
  1. Rubisco activase is an enzyme which regulates rubisco and uses ATP for the regulation of Rubisco
  2. In the rubisco-activase–mediated cycle, the hydrolysis of ATP by rubisco-activase elicits a conformational change of rubisco that reduces its binding affinity for sugar phosphates in presence of light
  3. Rubisco activase in some species is activated by light through a redox mechanism
  4. These light actuated ion fluxes decreases pH and increases that of Mg2+ by 2 to 5 mM, activate enzymes of the Calvin–Benson cycle that require Mg2+ for catalysis
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TLS Online TPP Program

#Question id: 14157

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

A ktup is composed of ___residues for protein sequences and _____ residues for DNA sequences.
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.