Maximum ratio of calcium, atp hydrolysis

[lilphy]

Hello,
1-The standard free energy associated to the hydrolysis of one ATP mole is -31 kJ. What would be the energy when [ATP]=5mM , [ADP]=0.5mM and [Pi]=1mM
2-Calculate the free energy required to move 1 mol of Ca2+ ions from the outside of the cell (where [Ca2+] = 10^-3) to the inside (where [Ca2+] = 10^-7) when the membrane potential is -70mV and the temperature is 25°C
3- What would be the maximum ratio of [Ca2+]out/[Ca2+]in that the atp hydrolysis would fight using the ATP ADP and PI concentrations and potential given above.

1/ I just used the formula
dG=dGo + RT ln ( [ADP][PI]/[ATP] )
I found dG=-53.8 kJ

2/ dG=RTln([Ca2+]o/[Ca2+]in) - 2*F*V
dG=36.3 kJ

3/ I calculated the equilibrium constants for each reaction, multiplied them to get [Ca2+]out[ADP][Pi] / [Ca2+]in[ATP] and then found the ratio from that is that correct ?
Thanks

 

[nate808]

for your post!

1/ Yes, your calculation for the free energy of ATP hydrolysis at those concentrations is correct. It is important to note that the standard free energy value of -31 kJ/mol is typically measured at standard conditions of 1M for all species involved, so the value may vary slightly at different concentrations.

2/ Your calculation for the free energy required to move Ca2+ ions across the membrane is also correct. This is known as the electrochemical potential and takes into account both the concentration gradient and the membrane potential.

3/ Yes, that is a correct approach for determining the maximum ratio of [Ca2+]out/[Ca2+]in that can be achieved through ATP hydrolysis. This ratio would depend on the concentrations of ATP, ADP, and Pi, as well as the membrane potential and temperature.