Practice Questions

  1. Home
  2. Free Resources
  3. Practice Questions
  4. 23806
TLS Online TPP Program

#Question id: 14796

Interception is a type of filter sterilization, which is used for the removal of 

#Part-B Specialized Branches in Biotechnology
  1. Particles smaller than the filter pore size
  2. Particles larger than the filter pore size
  3. Particles equal to the filter pore size
  4. All of the above
More Questions
TLS Online TPP Program

#Question id: 14160

#Part-B Specialized Branches 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: 14160

#Part-A Aptitude & General 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

#Part-B Specialized Branches in Biotechnology

Which of the following is not a description of dynamic programming algorithm?
TLS Online TPP Program

#Question id: 14161

#Part-A Aptitude & General Biotechnology

Which of the following is not a description of dynamic programming algorithm?
TLS Online TPP Program

#Question id: 14164

#Part-B Specialized Branches in Biotechnology

Stoichiometric equations are used to represent growth of microorganisms provided a ' molecular formula' for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, O and other elements in cells. For a particular bacterial strain, the molecular formula was determined to be C4.4H7.30 1.2N0.86 . These bacteria are grown under aerobic conditions with hexadecane (C16H34) as substrate. The reaction equation describing growth is:

                                  

  Assuming 100% conversion, what is the yield of cells from hexadecane in g g-l?   
TLS Online TPP Program

#Question id: 14165

#Part-B Specialized Branches in Biotechnology

Stoichiometric equations are used to represent growth of microorganisms provided a ' molecular formula' for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, O and other elements in cells. For a particular bacterial strain, the molecular formula was determined to be C4.4H7.30 1.2N0.86 . These bacteria are grown under aerobic conditions with hexadecane (C16H34) as substrate. The reaction equation describing growth is:

                                    

Assuming 100% conversion, what is the yield of cells from oxygen in g g-l?