What is the efficiency of oxidative phosphorylation from oxidation of 1 mole of NADH under standard conditions?
37%
50%
70%
78%
The correct answer and explanation is:
The correct answer is 37%.
Oxidative phosphorylation is the process by which cells produce ATP through the electron transport chain and chemiosmosis. This occurs in the mitochondria, where the energy from the oxidation of NADH and FADH2 is used to generate a proton gradient across the mitochondrial membrane. This proton gradient drives the production of ATP by ATP synthase.
The energy efficiency of oxidative phosphorylation depends on the amount of energy transferred from NADH to ATP. In standard conditions, the oxidation of 1 mole of NADH produces around 2.5 moles of ATP. The energy released during the oxidation of NADH is about 52.6 kcal/mol, while the production of ATP consumes approximately 7.3 kcal/mol of energy.
To calculate the efficiency, the energy yield from ATP production (2.5 moles × 7.3 kcal/mol = 18.25 kcal) is compared to the total energy released during NADH oxidation. This gives an efficiency of about 18.25 kcal / 52.6 kcal, which is approximately 37%.
This efficiency can vary depending on several factors, including the proton gradient, the mitochondrial membrane potential, and the coupling efficiency between the electron transport chain and ATP synthase. Under ideal conditions, the process is about 37% efficient, meaning that 37% of the energy from NADH oxidation is converted into usable ATP, while the rest is lost as heat.
Thus, oxidative phosphorylation is not a perfectly efficient process. A significant portion of the energy is lost as heat, which is why cellular respiration also produces heat, contributing to the regulation of body temperature in warm-blooded organisms.