How do the Krebs cycle and the electron transport chain complete the oxidation of glucose and generate most of the cell's ATP?
Describe the Krebs cycle and oxidative phosphorylation, including chemiosmosis and the role of oxygen as the final electron acceptor
A focused answer to the H2 Biology Energy and Equilibrium outcome on the final stages of aerobic respiration. The Krebs cycle in the matrix, the electron transport chain and chemiosmosis on the inner membrane, the role of oxygen, and the overall ATP yield.
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What this dot point is asking
SEAB wants you to describe the Krebs cycle (in the mitochondrial matrix) and oxidative phosphorylation (the electron transport chain and chemiosmosis on the inner membrane), to explain how most ATP is produced, and to state the role of oxygen as the final electron acceptor. This completes the story of aerobic respiration begun in glycolysis and the link reaction.
The answer
The Krebs cycle
The Krebs cycle occurs in the mitochondrial matrix. Each acetyl coenzyme A delivers its two-carbon acetyl group, which combines with a four-carbon molecule to form a six-carbon molecule (citrate); coenzyme A is released to be reused.
Through a series of reactions, the six-carbon molecule is converted back to the four-carbon acceptor. During this:
- two carbon dioxide are removed (decarboxylation),
- hydrogen is removed (dehydrogenation) and accepted to form three reduced NAD and one reduced FAD, and
- one ATP is made by substrate-level phosphorylation.
The four-carbon acceptor is regenerated, so the cycle continues. Because each glucose gives two acetyl coenzyme A, the cycle turns twice per glucose.
Oxidative phosphorylation
Most ATP is made here, on the inner mitochondrial membrane.
- Electron transport chain. Reduced NAD and reduced FAD deliver hydrogen, which splits into electrons and protons. The electrons pass along a chain of carrier proteins, releasing energy at each step.
- Proton pumping. This energy pumps protons from the matrix into the intermembrane space, building a proton gradient. This is chemiosmosis.
- ATP synthase. Protons flow back into the matrix through ATP synthase, and the energy of this flow drives the synthesis of ATP from ADP and inorganic phosphate.
- Oxygen. At the end of the chain, oxygen is the final electron acceptor, combining with electrons and protons to form water.
Overall yield
In total, the complete aerobic oxidation of one glucose yields about 30 to 32 ATP, the great majority from oxidative phosphorylation.
Examples in context
Example 1. Brown fat and uncoupling. In some tissues a protein lets protons leak back across the inner membrane without passing through ATP synthase, so the gradient's energy is released as heat instead of ATP. This shows that the proton gradient, not a direct chemical link, is what normally drives ATP synthesis.
Example 2. Reduced FAD enters lower down. Reduced FAD delivers its electrons further along the chain than reduced NAD, so fewer protons are pumped per reduced FAD and slightly less ATP results. This detail explains why the two coenzymes contribute different amounts to the total yield.
Try this
Q1. State the role of oxygen in oxidative phosphorylation. [1 mark]
- Cue. It is the final electron acceptor, combining with electrons and protons to form water.
Q2. Explain how a proton gradient is used to make ATP. [2 marks]
- Cue. Protons flow back into the matrix down their gradient through ATP synthase, and the energy of this flow drives the synthesis of ATP from ADP and inorganic phosphate.
Q3. State where in the mitochondrion the Krebs cycle and the electron transport chain are located. [2 marks]
- Cue. The Krebs cycle occurs in the matrix; the electron transport chain is on the inner membrane (the cristae).
Exam-style practice questions
Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
Original6 marksDescribe how the electron transport chain and chemiosmosis generate ATP during oxidative phosphorylation, and explain the role of oxygen.Show worked answer →
Examiners want the chain, the proton gradient, ATP synthase, and the role of oxygen.
Reduced NAD and reduced FAD deliver hydrogen to the electron transport chain on the inner mitochondrial membrane. The hydrogen splits into electrons and protons. The electrons pass along a series of carrier proteins, releasing energy at each transfer.
This energy is used to pump protons from the matrix into the intermembrane space, creating a proton gradient (a higher proton concentration outside the matrix). This is chemiosmosis.
The protons flow back into the matrix through the enzyme ATP synthase, and the energy of this flow drives the synthesis of ATP from ADP and inorganic phosphate.
Oxygen is the final electron acceptor: at the end of the chain it accepts the electrons and the protons to form water. Without oxygen the chain would back up, electrons could not be passed on, and ATP synthesis by this route would stop.
Markers reward the delivery of hydrogen by reduced coenzymes, electron transfer down the chain, proton pumping to form a gradient, ATP synthase using the return flow, and oxygen as the final electron acceptor forming water.
Original4 marksDescribe the main events and products of one turn of the Krebs cycle, starting from acetyl coenzyme A.Show worked answer →
The answer should cover the combination, the decarboxylations, and the reduced coenzymes.
Acetyl coenzyme A delivers its two-carbon acetyl group, which combines with a four-carbon molecule to form a six-carbon molecule (citrate). Coenzyme A is released to be reused.
The six-carbon molecule is then gradually converted back to the four-carbon molecule through a series of reactions. During these, two molecules of carbon dioxide are removed (decarboxylation), and hydrogen is removed (dehydrogenation) and accepted by coenzymes to form reduced NAD and reduced FAD. One ATP is also produced by substrate-level phosphorylation.
Per turn the products are 2 carbon dioxide, 3 reduced NAD, 1 reduced FAD, and 1 ATP, and the four-carbon molecule is regenerated to accept the next acetyl group.
Markers reward the combination of the acetyl group with the four-carbon molecule to form citrate, the two decarboxylations, the reduced coenzymes, the ATP, and the regeneration of the acceptor molecule.
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