TLS Online TPP Program

#Question id: 21003


In India, a consortium of bacterial species has been developed to combat oil spills and oily sludge; the inoculant is aptly called 

#Unit 12. Applied Biology
  1. Oilcalciterant
  2. Oilzapper
  3. Oil degrading
  4. Bioremediation
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TLS Online TPP Program

#Question id: 2235

#Unit 2. Cellular Organization

The generation and maintenance of a membrane electric potential requires which of the following:

a. active pumping of ions       

b. ion-specific membrane channel proteins

c. ATP                                    

d. a generally permeable membrane

TLS Online TPP Program

#Question id: 2236

#Unit 2. Cellular Organization

The partition coefficient K, the equilibrium constant for partitioning between oil and water, for butyric acid is about 10–2, and for 1,4-butanediol it is about 10–4.  You add liposomes containing only water to a solution with an initial concentration of 1 mM butyric acid and 100 mM 1,4-butanediol outside the liposomes.  What is the relative rate of diffusion of the two substances into the liposome interior?

TLS Online TPP Program

#Question id: 2237

#Unit 2. Cellular Organization

Glucose enters erythrocytes via a GLUT-1 uniporter. As the levels of glucose in the bloodstream decrease between meals, what happens to the glucose in the cells?

TLS Online TPP Program

#Question id: 2238

#Unit 2. Cellular Organization

Consider the transport of glucose into an erythrocyte by facilitated diffusion.  When the glucose concentrations are 5 mM on the outside and 0.1 mM on the inside, the free-energy change for glucose uptake into the cell is:  (These values may be of use to you:  R = 8.315 J/mol·K; T = 298 K; 9 (Faraday constant) = 96,480 J/V; N = 6.022×10²³ /mol.)

TLS Online TPP Program

#Question id: 2239

#Unit 2. Cellular Organization

Consider the transport of K+ from the blood (where its concentration is about 4 mM) into an erythrocyte that contains 150 mM K+.  The transmembrane potential is about 60 mV, inside negative relative to outside.  The free-energy change for this transport process is:  (These values may be of use to you:  R = 8.315 J/mol.K; T = 298 K; 9 (Faraday constant) = 96,480 J/V; N = 6.022×10²³ /mol.)

TLS Online TPP Program

#Question id: 2240

#Unit 2. Cellular Organization

 In which of the following cases is energy NOT needed for transmembrane transport?

a. Lysine moves into the cell against its concentration gradient via the Na+/lysine symporter.

b. Potassium ions (K+) move out of the cell down the K+ concentration gradient via potassium channels.

c. Glucose moves into the cell down its concentration gradient via a glucose uniporter.

d. Sodium ions (Na+) move out of the cell down the Na+ concentration gradient via sodium channels.

Which of the following are correct?