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

The efficiency of ATP synthesis in cells from the oxidation of NADH is approximately 56%, calculated by comparing the energy yield per NADH during ATP synthesis to the energy released during NADH oxidation.

Step-by-step explanation:

The efficiency of ATP synthesis in cells can be calculated by comparing the standard-state free energy change during the oxidation of NADH to the energy cost of synthesizing ATP.

Given that oxidation of NADH with oxygen is -220 kJ/mol (-52.6 kcal/mol) and that the synthesis of one mole of ATP requires 11.7 kcal/mol, for every NADH, 2.5 ATP are synthesized, resulting in a total energy yield of 2.5 ATP × 11.7 kcal/mol = 29.25 kcal/mol of NADH.

To find the efficiency, we take the energy yield per mole of NADH and divide it by the energy released per mole of NADH when oxidized by oxygen, and then multiply by 100 to convert to a percentage.

The calculation becomes:

Efficiency (%) = (Energy yield per NADH / Energy released by oxidation of NADH) × 100
Efficiency (%) = (29.25 kcal/mol / 52.6 kcal/mol) × 100
Efficiency (%) = 55.6%

To the nearest ones, the efficiency of ATP synthesis from the oxidation of NADH is approximately 56%.