During the process of cellular respiration, oxygen \((O_2)\) plays a critical role, especially in the final stage called oxidative phosphorylation within the mitochondria. To understand how oxygen is used, we'll analyze the options given:
A. To combine with the depleted energy carrier \(NAD^+\) and form \(NADH\), which carries energy
This statement is incorrect. \(NAD^+\) is reduced to form \(NADH\) earlier in cellular respiration during glycolysis and the Krebs cycle. This process does not involve oxygen directly. \(NADH\) is an electron carrier that later donates electrons to the electron transport chain.
B. To combine with the 4-C starting material of the Krebs cycle and form a 6-C molecule
This option is also incorrect. The 4-carbon molecule in the Krebs cycle is oxaloacetate, which combines with the 2-carbon molecule acetyl-CoA to form the 6-carbon molecule citrate. Oxygen does not participate in this reaction.
C. To provide the energy that is needed to cause chemical bonds in glucose to break
This statement is incorrect as well. The energy required to break bonds in glucose comes from the investment of ATP during glycolysis. Oxygen does not directly provide this energy. Instead, the breakdown of glucose generates high-energy molecules such as NADH and FADH2, which later use oxygen in the electron transport chain.
D. To accept electrons \((e^-)\) at the end of an electron transport chain and cause water \((H_2O)\) to form
This statement is correct. In the final stage of cellular respiration, known as oxidative phosphorylation, electrons are transferred through a series of protein complexes in the electron transport chain. At the end of the chain, oxygen acts as the final electron acceptor. It combines with electrons and hydrogen ions to form water \((H_2O)\). This step is crucial for the production of ATP, as it maintains the flow of electrons and facilitates the generation of the proton gradient required for ATP synthesis.
Therefore, the correct answer is:
D. To accept electrons [tex]\((e^-)\)[/tex] at the end of an electron transport chain and cause water [tex]\((H_2O)\)[/tex] to form
