1.) Glucose 6-Phosphate + H2O -> Glucose + Pi has a Keq of 270. ATP + Glucose ->
ID: 51025 • Letter: 1
Question
1.) Glucose 6-Phosphate + H2O -> Glucose + Pi has a Keq of 270. ATP + Glucose -> ADP + Glucose 6 phosphate has a Keq of 8 9 0. Using this information, calculate the standard free energy of hydrolysis of ATP.
2.)The complete combustion of glucose to CO2 and H2O proceeds with an overall G° of -686 kcal/mol. When this process occurs in a typical cell, 36 mol of ATP are produced from ADP and Pi.
Assuming that the G for ATP <--> ADP + Pi is -10 kcal/mol, and that G° = G for glucose oxidation under cellular conditions, what fraction of the potential energy of glucose is conserved as chemical bond energy in ATP?
What happens to the energy not conserved as ATP bond energy?
Explanation / Answer
1. caluculation of free energy of hydrolysis of ATP;
Go ’= -RT ln K’eq , R =-8.315 J / mol , T =298 , so the delta G for the above reactions should be caluculated ;
for reaction 1 ; Go ’ = - 8.315 *298 ln 270 = -13.9 Kj/mol
for reaction 2 ; Go ’= -8.315 *298 ln 890 = - 16.8kj/mol
so the free energy for the ATp is -13.9+ [-16.8] = -30.7 kj
2. the complete oxidation of the combustion of glucose is 36 ATP molecules include the production of ATP from the mitochondria . this type of free energy occurs when cells are at room temperature and cellular respiration ,this free nergy change occurs in the cells by the break of C-bond of six carbon atom , the negative indicates the products produce less free energy than the reactants. potential energy is stored in the form of chemichal bonds connecting the atoms in the molecules . it is not actually the breaking of atp does not produce . The breakage of bonds ATP hydrolysis reaction so if the cell do not store energy in the form the ATP the cells cannot produce the free nergy for the metabolism
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.