Question

Complete the following calculations. Include units and work for full points. A. Calculate the actual free energy of hydrolysis of ATP, AG, in a E. coli Cell. The concentration is ATP, ADP and Pi, are shown in Table 13-5 shown. TABLE 13-5 Total Concentrations of Adenine Nucleotides, Inorganic Phosphate, and Phosphocreatine in Some Cells Concentration (mm) ATP ADPD AMPP, PC Rat hepatocyte 3.38 1.32 0.29 4.8 0 Rat myocyte 8.05 0.93 0.04 8.05 28 Rat neuron 2.59 0.73 0.06 2.72 4.7 Human erythrocyte 2.25 0.25 0.02 1.65 0 E. coli cell 7.90 1.04 0.82 7. 9 0 For erythrocytes the concentrations are those of the cytosol (human erythrocytes lack a nucleus and mitochondria). In the other types of cells the data are for the entire cell contents, although the cytosol and the mitochondria have very different concentrations of ADP.PCris phosphocreatine, discussed on p. 516. This value reflects total concentration, the true value for free ADP may be much lower (p. 509).

Can anyone help me with these calculations? Thanks!

Answer 1

Actual face energy of hydrolysis of ATP AG = AGO + RT la Kag - In ATP hydrolysis, ATP + H₂O Ž ADP + Po Kea = [ADP] [ſ ] (ATP] At equilibrium , AG = 0; DG = AG+RTIM ADP] [ ] [ATP] I Elibbs foree energy undes standard condition varies from – 28 to -34 kJ/mol So for convenience, we take in AGi = 30.5 kJ/mol (.. It is a constant) SOP] [P. - ③

from given values, [App] = 1.0A X103M (: they are expressed [p] = 7.9x103 as molar values) [ATP] = 7.90x103 We know, R=8.315/mol T = 298 K Putting all nalues in equation , (.DAX 10 x 10 ) AG = 30.5 + 8.31x298 x 2.503x logo (64.90x10-3) AG = 30.5 +8.51x298x 2.30% AG = – 1688kJ molt So the actual free energy of hydrolysis of AT in E. coli cell is 46.981 5/mol.