Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Cell respiration is the controlled release of energy from organic compounds to produce . It involves metabolic pathways including glycolysis, the link reaction, the Krebs cycle, and oxidative phosphorylation.
Glycolysis occurs in the cytosol and is anaerobic. It involves the phosphorylation of glucose, lysis into triose phosphate, and oxidation to produce molecules of pyruvate, with a net yield of and .
In the Link Reaction, pyruvate is transported into the mitochondrial matrix, where it undergoes decarboxylation and oxidation to form an acetyl group, which binds to Coenzyme A to form .
The Krebs Cycle occurs in the matrix. Each turn processes one , releasing and generating , , and via substrate-level phosphorylation.
The Electron Transport Chain (ETC) is located on the inner mitochondrial membrane (cristae). High-energy electrons from and are passed through carriers, releasing energy used to pump protons () into the intermembrane space.
Chemiosmosis is the diffusion of protons down their electrochemical gradient from the intermembrane space back into the matrix through synthase, which catalyzes the synthesis of .
Oxygen () is the final electron acceptor in the . It combines with electrons and ions to form water (), maintaining the proton gradient by removing de-energized electrons.
Mitochondrial structure is adapted to function: the cristae provide a large surface area for the , and the small intermembrane space allows for rapid accumulation of protons to create a concentration gradient.
📐Formulae
💡Examples
Problem 1:
Explain why the yield of from one molecule of is lower than the yield from one molecule of .
Solution:
yields approximately to , while yields approximately to .
Explanation:
donates its electrons to Complex I of the Electron Transport Chain, whereas donates its electrons further down the chain at Complex II. Consequently, the electrons from trigger the pumping of fewer protons () across the inner membrane, resulting in a smaller proton motive force and less produced via chemiosmosis.
Problem 2:
Determine the net production of and during the Krebs cycle for a single molecule of glucose.
Solution:
and .
Explanation:
One glucose molecule produces two pyruvate molecules, which lead to two turns of the Krebs cycle. Each turn of the cycle produces and . Therefore, and .