TLS Online TPP Program

#Question id: 4586


 What are adaptations?

#SCPH06 I Botany
  1. geologic changes over time

  2. rocks containing fossils

  3. inherited characteristics of organisms that enhance their survival

  4. descent with modification from a common ancestor

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TLS Online TPP Program

#Question id: 14161

#I Life Science/ Life Sciences Group – I-V

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

TLS Online TPP Program

#Question id: 14161

#SCPH05 I Biotechnology

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

TLS Online TPP Program

#Question id: 14161

#SCPH05 I Biotechnology

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

TLS Online TPP Program

#Question id: 14162

#SCPH05 I 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.

TLS Online TPP Program

#Question id: 14163

#SCPH05 I Biotechnology

A stream of growth medium containing 60% glucose and 40% growth factor A flowing at 500 kg per hour was divided into two feed streams for different cell cultures. One fourth of the growth medium was sent to fermenter 1 and the rest was sent to fermenter 2. What is the flow rates of glucose and growth factor A in the feed streams for the two fermenters

TLS Online TPP Program

#Question id: 14164

#SCPH05 I 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?