TLS Online TPP Program

#Question id: 15701


In a sequence logo of the type given below, the sizes of the letters are proportional to the:

#I Life Science/ Life Sciences Group – I-V
  1. Number of residues in the sequences
  2. Information content of the respective residues 
  3. Frequencies of the respective residues in the sequences 
  4. Resolution of the output device (terminal/printer) 
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TLS Online TPP Program

#Question id: 70

#SCPH05 I Biotechnology

The dimensions of living cells are limited, on the lower end by the minimum number of biomolecules necessary for function, and on the upper end by the rate of diffusion of solutes such as oxygen. Except for highly elongated cells, they usually have lengths and diameters in the range of

TLS Online TPP Program

#Question id: 70

#SCPH06 I Botany

The dimensions of living cells are limited, on the lower end by the minimum number of biomolecules necessary for function, and on the upper end by the rate of diffusion of solutes such as oxygen. Except for highly elongated cells, they usually have lengths and diameters in the range of

TLS Online TPP Program

#Question id: 71

#SCPH05 I Biotechnology

Which of the following is not the priority rule for R, S Configuration?

TLS Online TPP Program

#Question id: 71

#SCPH06 I Botany

Which of the following is not the priority rule for R, S Configuration?

TLS Online TPP Program

#Question id: 72

#SCPH05 I Biotechnology

For the reaction catalyzed by the enzyme hexokinase: Glucose + ATP → glucose 6-phosphate + ADP the equilibrium constant, Keq, is 7.8 × 102. In living E. coli cells, [ATP] = 5 mM, [ADP] = 0.5 mM, [glucose] = 2 mM, and [glucose 6-phosphate] = 1 mM. Which of the following conclusions made on the basis of the above calculations is incorrect?

TLS Online TPP Program

#Question id: 72

#SCPH06 I Botany

For the reaction catalyzed by the enzyme hexokinase: Glucose + ATP → glucose 6-phosphate + ADP the equilibrium constant, Keq, is 7.8 × 102. In living E. coli cells, [ATP] = 5 mM, [ADP] = 0.5 mM, [glucose] = 2 mM, and [glucose 6-phosphate] = 1 mM. Which of the following conclusions made on the basis of the above calculations is incorrect?