Suppose the concentration of glucose inside a cell is 0 1 mM and the cell is sus

Question

Suppose the concentration of glucose inside a cell is 0 1 mM and the cell is suspended in a glucose solution of 0.01 nM What would be the free energy change involved in transporting 10 -6 mole of glucose from the medium into the cell? Assume T = 37 degree C. What would be the free energy change involved in transporting 10 -6 mole of glucose from the medium into the cell if the intercellular and extracellular concentrations were 1 nM and 10 mM, respectively? If the processes described in parts A and B were coupled to ATP hydrolysis, how many moles of ATP would have to be hydrolyzed in order to make each process favorable? (Use the standard free energy change for ATP hydrolysis )

Explanation / Answer

M1 = 0.1 M

M2 = 0.01 M

n = 10^-6 mol to the cell

T = 37C or 310K

a)

By definition

dG = R*T*ln([in]/[out]) = 0.082*(310)*(0.1/0.01)= 254.2 J/mol

But this is per unit mol, we hav e10^-6 mol so:

254.2 * (10^-6) = 0.000254 J

b)

Apply same formula

dG = R*T*ln([in]/[out]) = 0.082*(310)*(1/10)= -2.54 J/mol

But we have 10^-6 mol so

-0.0000254 J

c)

Total reactions:

ATP+H2OADP+Pi      G = -30.5 kJ/mol

ATP+H2OAMP+2Pi         G = -61 kJ/mol

2ADP+H2O2AMP+2Pi      G = -61 kJ/mol

Total for 1 mol of ATP = -30.5 -61 - 61 = -152.5 kJ/mol

If we have

When you couple them

dG = dG1 + dG2 = 254.2 - 2.54 = 251.66 J in total

therefore

dG/n = 251.66 J / 152500 J/mol = 0.00165 mol or 1.65 mmol

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