Calculate the free energy of hydrolysis of ATP (to ADP and Pi) in a ral liver ce

Question

Calculate the free energy of hydrolysis of ATP (to ADP and Pi) in a ral liver cell at 37 degree C in which the ATP. ADP. and Pi concentrations are 5.6, 9.3. and 4.1 mM, respectively. Express your answer in kJ to the nearest tenth. Fructose-l,6-bisphosphate (FBP) glyceraldehyde-3-phosphate (GAP) + dihydroxyacetone phosphate (DHAP) The standard free energy change for this reaction is +22.8 kJ/mol. Under actual conditions in the cell the concentrations of GAP, DHAP, and FBP are 0.4 mM, 0,2 mM, and 4.0 mM respectively. Given this information calculate the actual free energy change for this reaction in the cell at a temperature of 37 degree C. Provide your answer (just the numerical part) in kJ/mole to two signicant figures. Example: 3.1

Explanation / Answer

2) The reaction can be given as

ATP -------> ADP + Pi

K= (ADP) (Pi) / (ATP)

Given (ADP)= 9.3 ,(Pi)=4.1 ,(ATP) =5.6

K= 9.3 x 4.1 /5.6

K= 6.81

Given Temperature T= 37 C = 37 + 273 = 310 K

R= 8.314

We know that dG= dGo + RT ln K

dGo= -30.5 kJ

RTln K = 8.314 x 310 ln 6.81

= 4944.35 J

RT lnK = 4.94 kJ

Free energy of Hydrolysis dG = -30.5 + 4.94

dG = -25.56 kJ

Free energy of hydrolysis is -25.56 kJ

3)

The reaction is given by

FBP ---> GAP + DHAP

K= (GAP) (DHAP)/(FBP)

Given (GAP) = 0.4 (DHAP) = 0.2 ,(FBP) = 4

K= 0.4 x 0.2 /4

K= 0.02

Given standard free energy change dGo = 22.8 kJ/mol

Temp T =37 C = 310 K

we know that dG= dGo+ RTlnK

RTlnK = 8.314 x 310 ln 0.02

RTlnK = -10.08 kJ/mol

dG= 22.8 - 10.08

dG= 12.72 kJ/mol

standard free energy is 12.72 kJ/mol

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