TLS Online TPP Program

#Question id: 13092


To express a yeast gene in E. coli, your task is to design a strategy to insert the yeast gene into the bacterial plasmid. Below is a map of the area of the yeast genome surrounding the gene in which you are interested.
 
The distance between each tick mark placed on the line above is 100 bases in length
Below are the enzymes you can use, with their specific cut sites shown 5’-XXXXXX-3’ 3’-XXXXXX-5’
 
The plasmid is 5,000 bases long and the two farthest restriction enzyme sites are 200 bases apart. The plasmid has an ampicillin resistance gene somewhere on the plasmid distal from the restriction cut sites.
                              
Which two restriction enzymes would you use to design a way to get the insert into the vector if you had to use two enzymes simultaneously?

#Section 7: Recombinant DNA technology and Other Tools in Biotechnology
  1. N (Nde1) and  S (sal1)
  2. EcoR 1 and K (Kpn 1)
  3. X (XbaI) and N (Nde1)
  4. K (KpnI) and X (XbaI)
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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.

TLS Online TPP Program

#Question id: 14163

#Section 4: Fundamentals of Biological Engineering

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: 14163

#Section 5: Bioprocess Engineering and Process 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

#Section 4: Fundamentals of Biological Engineering

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: 14164

#Section 5: Bioprocess Engineering and Process 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

#Section 4: Fundamentals of Biological Engineering

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?