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Solutions4 - BC368 Biochemistry of the Cell II Problem...

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BC368 – Biochemistry of the Cell II Problem Set #4 – Overview of Metabolism ANSWERS 1. Because the concentrations of reactants and products deviate from the standard 1 M, the standard free energy of hydrolysis (-30.5 kJ/mol) does not apply. G= G ’O -RT lnQ For ATP hydrolysis: ATP + H 2 O <--> ADP + Pi; Q = [ADP][Pi]/[ATP] G(liver)= -30.5 kJ/mol + (8.315 J/K-mol) x (310K) x (ln {(1.8 x 10 -3 )(5.0 x 10 -3 )/(3.5 x 10 -3 )} G(liver)= -45.9 kJ/mol G(muscle)= -30.5 kJ/mol + (8.315 J/K-mol) x (310K) x (ln {(0.9 x 10 -3 )(8.0 x 10 -3 )/(8.0 x 10 -3 )} G(muscle)= -48.6 kJ/mol G(brain)= -30.5 kJ/mol + (8.315 J/K-mol) x (310K) x (ln {(2.7 x 10 -3 )(0.7 x 10 -3 )/(2.6 x 10 -3 )} G(brain)= -49.1 kJ/mol MOST FAVORABLE. 2. a) As Mg 2+ concentration falls, G becomes more negative (more favorable). b) Mg 2+ binds to the phosphates of ATP and helps to alleviate the charge repulsion. When Mg 2+ concentration decreases, there is an increase in charge repulsion and the reaction becomes more favorable. 3. Energetically the two molecules are equivalent. However, if your assay is an enzyme
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