TLS Online TPP Program

#Question id: 4954


If natural selection in a particular environment favored genetic systems that permitted the production of daughter ʺcellsʺ that were genetically dissimilar from the mother ʺcells,ʺ then one should expect selection for

I. polynucleotide polymerase with low mismatch error rates.

II. polynucleotide polymerases without proofreading capability.

III. batteries of efficient polynucleotide repair enzymes.

IV. polynucleotide polymerases with proofreading capability.

V. polynucleotide polymerases with high mismatch error rates.

#SCPH06 I Botany
  1. I only

  2. I and IV

  3. I, III, and IV

  4. II and V

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TLS Online TPP Program

#Question id: 14266

#SCPH05 I Biotechnology

For the activated-sludge, the specific growth rate of cells is given by 


 

The following parameter values are known: F = 500 l/h, α = 0.4, ɣ = 0.1, Xe = 0, V = 1500 l, Ks = 10 mg/l, µm = 1 h-1 , kd = 0.05 h^-1 , S0 = 1000 mg/l, Y M X/S = 0.5 g dw/g substrate.. Calculate the cell concentration in the reactor. 

TLS Online TPP Program

#Question id: 14267

#SCPH05 I Biotechnology

In a trickling biological filter, the BOD value of the feed stream is S0i = 500 mg/l with a feed flow of F = 103 l/h. The effluent BOD value is desired to be S0 = 10 mg/l. The following kinetic parameters for the biocatalysts are known: rm = 20 mg/S/l .h and Ks = 200 mg S/l. The biofilm thickness is L = 0.1 mm. The cross-sectional area of the filter is A = 2 m2 , and the biofilm surface area per unit volume of the bed is a = 500 cm2 /cm3 . Assume that dissolved oxygen is the rate-limiting substrate and the diffusion coefficient of oxygen is DO2 = 2 × 10-5 cm2 /s. Determine the required height of the bed. You can assume first-order bioreaction kinetics?

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TLS Online TPP Program

#Question id: 14268

#SCPH05 I Biotechnology

An activated-sludge waste treatment system is required to reduce the amount of BOD5 from 1000 mg/l to 20 mg/l at the exit. The sedimentation unit concentrates biomass by a factor of 3. Kinetic parameters are µm = 0.2 h-1 , Ks = 80 mg/l, kd = 0.01 h-1 , and Y M X/S = 0.5 g MLVSS/g BOD5. The flow of waste water is 10000 l/h and the size of the treatment basin is 50,000 l. What is the value of the solids residence time?

TLS Online TPP Program

#Question id: 14269

#SCPH05 I Biotechnology

An activated-sludge waste treatment system is required to reduce the amount of BOD5 from 1000 mg/l to 20 mg/l at the exit. The sedimentation unit concentrates biomass by a factor of 3. Kinetic parameters are µm = 0.2 h-1 , Ks = 80 mg/l, kd = 0.01 h-1 , and Y M X/S = 0.5 g MLVSS/g BOD5. The flow of waste water is 10000 l/h and the size of the treatme nt basin is 50,000 l. What value of the recycle ratio must be used?

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TLS Online TPP Program

#Question id: 14271

#SCPH05 I Biotechnology

Penicillin is produced by P. chrysogenum in a fed-batch culture with the intermittent addition of glucose solution to the culture medium. The initial culture volume at quasi-steady state is V0 = 500 l, and glucose-containing nutrient solution is added with a flow rate of F = 50 l/h. Glucose concentration in the feed solution and initial cell concentration are S0 = 300 g/l and X0 = 20 g/l, respectively. The kinetic and yield coefficients of the organism are mm = 0.2 h-1 , KS = 0.5 g/l, and YX/S = 0.3 g dw/g glucose. Determine the concentration of cells at quasi-steady state when t = 10 h.