Please provide step by step calculations. What is the free energy change if 1 mo

Question

Please provide step by step calculations.

What is the free energy change if 1 mole of Na+ is transported across a membrane from a region where the concentration is 1 ?M to a region where it is 100 ?M? Assume T = 37oC. (It will probably help you to make a sketchof this system before attempting calculations.)

A. In the absence of a membrane potential.
B. When the transport is opposed by a membrane potential of 70 mV.
C. In each case, will hydrolysis of 1 mole of ATP suffice to drive the transport of 1 mole

of ion? (Assume ?G for ATP hydrolysis is -50 kJ/mol under these conditions.)
D. The average human (70 Kg) generates 40 Kg of ATP every 24 hrs. If 50% of this ATP is used in

ion transport, mainly by Na+/K+ ATPase, calculate how many grams of Na+ and K+ the average human pumps across membranes every 24 hrs.

R = 8.3 J M-1 oK-1, F = 96500 J mol-1 v-1
Mol. weight of ATP = 503 g/mol
Atomic weight of Na = 23 g/mol and of K = 39 g/mol

Explanation / Answer

Answer:

The complete balanced reaction is

3 Na+(in) + 2 K+(out) + ATP ------ > 3 Na+(out) + 2 K+(in) + ADP + Pi

Hence the G for the above reaction is

G(Na pump) =  G(ATP hydrolysis) + G(Na+ transprt) +  G(k+ transport)  

Given   G(ATP hydrolysis) =  -50 kJ/mole

(A) In absence of any membrane potential,Na+ (in) = K+(out).

Hence  G(Na+ transprt) = G(k+ transport) = 0

Hence  G(Na pump) =  G(ATP hydrolysis) +  G(Na+ transprt) +  G(k+ transport)  

=  - 50 kJ/mole + 0 + 0

=  - 50 kJ/mole (answer)

Since G(Na pump) is negative, it is suffucient to carry out the reaction

(B) Now the transport is opposed by a membrane potential of 70 mV = 0.070 V

given [Na+](in) = 1x10-6 M

[Na+](out) = 100x10-6 M

T = 37 DegC = 37 + 273 = 310 K

In soium potassium pump, for every 3 Na+ out, 2 K+ are in.

In this situatiion 2 types of work need to be done

1. Chemical work of pumping the ion against the membrane potential

2. Electrical work of pumping the ion against the membrne potential.

The above work done will be equal to the Gibbs free energy change.

Since in soium potassium pump, for every 3 Na+ out, 2 K+ are in,

G(chemical work) = 3 x nRTxln[Na+](out) / [Na+](in) = 3 x 1 x 8.314 x 310K x ln( 100x10-6 M /  1x10-6 M)

= 35607 J/mol = 35.61 KJ/mol

G(electrical work) = 3ZFfi = - 3x1 x (96.5 KJ/molxV) x ( - 0.070 V) = + 20.27 KJ/mol

G(total, Na+) = G(chemical work) + G(electrical work) = 35.61 KJ/mol + 20.27 KJ/mol

= 56.0 KJ/mol

Similarly G(total, K+) =    G(chemical work) + G(electrical work)

= 2 x nRTxln[K+](out) / [K+](in) + 2ZFfi

= 2 x 1 x 8.314 x 310K x ln( 100x10-6 M /  1x10-6 M) + 2x1 x (96.5 KJ/molxV) x ( - 0.070 V)

= + 17 KJ/mol

Hence

G = G(Na pump at 0 membrane potential) + G(total, Na+) + G(total, K+)

= - 50 KJ/mol +  56.0 KJ/mol +  17 KJ/mol

= + 23 KJ/mol (answer)

Since G is positive,the reaction is not spontaneous and hence the ATP is not sufficient.

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