Answered: In considering active transport by Na +…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Question

In considering active transport by Na + -K + -ATPase at body temperature (37 o C), 3 Na+ are pumped out of the cell and 2 K + are pumped in for each ATP that is hydrolyzed to ADP + P i . Given that underyour experimental conditions, the DG for ATP hydrolysis is -10 kcal/mol, and that V is -60 mV, and that the pump maintains the internal Na + at 10mM, external Na + at 120 mM, internal K + at 120 mM and external K + at 8mM, what is the efficiency of the pump (i.e., what fraction of the energy available from ATP hydrolysis is required to drive transport at the provided levels)?

Expert Solution

Step 1: Free energy change for transport of solutes

The following equation describes the mathematical relation for the change in free energy $open parentheses increment G close parentheses$ required to transport a solute against its concentration gradient:

$increment G space equals space R times T times ln space open parentheses fraction numerator C 2 over denominator C 1 end fraction close parentheses$

where:

R is the gas constant 8.314 J mol^-1 K^-1
T is the temperature in Kelvin
C2 is the ion concentration of where the ion is going
C1 is the ion concentration of where the ion is coming from

Now, if the solute is an ion, then we must factor in not only the change in chemical gradient (described above) but also the electrical gradient. The above equation changes to:

$increment G space equals space R times T times ln space open parentheses fraction numerator C 2 over denominator C 1 end fraction close parentheses space plus space Z times calligraphic F times increment E$

where:

Z is the charge on the ion
ΔE is the transmembrane potential
$calligraphic F$ is the faraday constant = 96480 J V^-1 mol^-1

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