The outer plasma membrane of a human cell is a 7.00-nm thick insulating layer. F

Question

The outer plasma membrane of a human cell is a 7.00-nm thick insulating layer. For a typical human cell of radius 5.00 m, this plasma membrane covers a surface area of 314 m2. The interior and exterior of the cell are conductive aqueous solutions containing lots of different types of ions, with the cell maintaining a charge imbalance between its interior and exterior using a variety of ATP-coupled ion pumps. Suppose a give cell maintains an ion imbalance such that its interior has a net negative charge of -1050000e (and its exterior has a similar net positive charge). A) What is the potential difference (inside - outside) across this cell's plasma membrane? Don't forget that the dielectric constants of water and the plasma membrane are 8.0 x 10 and 5.5, respectively Number 7.75x 10-8 B) The major contributor to the cell's ion imbalance is the Na K exchange pump. This pump hydrolyzes one ATP molecule (ATP-> ADP + P) and uses that chemical energy to pump 3 Na* ions out of the cell and 2 K ions into the cell. What is the net change in electric potential energy for each 3-Na*-for-2-K exchange? (Hint: For reasonableness, compare your answer to the energy released by a single ATP hydrolysis.) Number Incorrect.

Explanation / Answer

Ans. As per the equation the electric charge would be

q = Ax P.E.x Ex E0/dxe

Where q = charge across the membrane

A = Surface area

P.E. = Potential difference

E = dielectric constant of membrane

E0 = vacuum permittivity

d = diameter = 2r

e = electron charge = 1.6 x 10-19

- 1050000 = 314 x 10-12 x P.E. x 5 x 8.8 x 10-12 / 10 x 10-6 x 1.6 x 10-19

Therefore P.E. = -1050000 x 10 x 1.6 x 10-25/ 314 x 10-12 x 5 x 8.8 x 10-12

Hence the potential difference across the cell membrane would be 121.59 V

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