Question

The most recent estimates for the stoichiometry of ATP synthesis from the oxidation of NADH are 2.5 ATP/NADH. Given that the standard-state free energy change in oxidizing NADH with oxygen is -220 kJ/mol (-52.6 kcal/mol), then what is the efficiency, expressed as a percent to the nearest ones, of ATP synthesis in cells if the cost of ATP synthesis is 11.7 kcal/mol?

Answer 1

Since one NADH generates 2.5 molecules of ATP with each molecule containing about 11.7 kcal of energy, the total amount of energy produced will be

2.5 * 11.7 kcal = 29.25 kcal

Since NADH contains 52.6 kcal of energy but only generates 29.25 kcal of energy

The efficiency, expressed as a percent to the nearest ones

The efficiency = (Amount of Kcal in NADH generated into ATP / amount of kcal within NADH) * 100

= (29.25 / 52.6) * 100 = 55.6%

= 56% (rounded to the nearest ones)