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Metabolism, Cell Respiration and Photosynthesis (AHL) - Cell Respiration (HL)

Grade 12IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Cell respiration is the controlled release of energy from organic compounds to produce ATPATP. It involves metabolic pathways including glycolysis, the link reaction, the Krebs cycle, and oxidative phosphorylation.

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Glycolysis occurs in the cytosol and is anaerobic. It involves the phosphorylation of glucose, lysis into triose phosphate, and oxidation to produce 22 molecules of pyruvate, with a net yield of 2 ATP2\text{ ATP} and 2 NADH+H+2\text{ NADH} + H^+.

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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 Acetyl-CoAAcetyl\text{-}CoA.

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The Krebs Cycle occurs in the matrix. Each turn processes one Acetyl-CoAAcetyl\text{-}CoA, releasing 2 CO22\text{ CO}_2 and generating 3 NADH+H+3\text{ NADH} + H^+, 1 FADH21\text{ FADH}_2, and 1 ATP1\text{ ATP} via substrate-level phosphorylation.

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The Electron Transport Chain (ETC) is located on the inner mitochondrial membrane (cristae). High-energy electrons from NADHNADH and FADH2FADH_2 are passed through carriers, releasing energy used to pump protons (H+H^+) into the intermembrane space.

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Chemiosmosis is the diffusion of protons down their electrochemical gradient from the intermembrane space back into the matrix through ATPATP synthase, which catalyzes the synthesis of ATPATP.

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Oxygen (O2O_2) is the final electron acceptor in the ETCETC. It combines with electrons and H+H^+ ions to form water (H2OH_2O), maintaining the proton gradient by removing de-energized electrons.

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Mitochondrial structure is adapted to function: the cristae provide a large surface area for the ETCETC, and the small intermembrane space allows for rapid accumulation of protons to create a concentration gradient.

📐Formulae

C6H12O6+6O2→6CO2+6H2O+Energy (ATP)C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy (ATP)}

Glycolysis: Glucose+2ADP+2Pi+2NAD+→2 Pyruvate+2ATP+2NADH+2H+\text{Glycolysis: } \text{Glucose} + 2ADP + 2P_i + 2NAD^+ \rightarrow 2\text{ Pyruvate} + 2ATP + 2NADH + 2H^+

Link Reaction: Pyruvate+NAD++CoA-SH→Acetyl-CoA+CO2+NADH+H+\text{Link Reaction: } \text{Pyruvate} + NAD^+ + \text{CoA-SH} \rightarrow \text{Acetyl-CoA} + CO_2 + NADH + H^+

Terminal Electron Acceptance: 12O2+2e−+2H+→H2O\text{Terminal Electron Acceptance: } \frac{1}{2}O_2 + 2e^- + 2H^+ \rightarrow H_2O

💡Examples

Problem 1:

Explain why the yield of ATPATP from one molecule of FADH2FADH_2 is lower than the yield from one molecule of NADHNADH.

Solution:

NADHNADH yields approximately 2.52.5 to 3 ATP3\text{ ATP}, while FADH2FADH_2 yields approximately 1.51.5 to 2 ATP2\text{ ATP}.

Explanation:

NADHNADH donates its electrons to Complex I of the Electron Transport Chain, whereas FADH2FADH_2 donates its electrons further down the chain at Complex II. Consequently, the electrons from FADH2FADH_2 trigger the pumping of fewer protons (H+H^+) across the inner membrane, resulting in a smaller proton motive force and less ATPATP produced via chemiosmosis.

Problem 2:

Determine the net production of CO2CO_2 and NADHNADH during the Krebs cycle for a single molecule of glucose.

Solution:

4 molecules of CO24\text{ molecules of } CO_2 and 6 molecules of NADH6\text{ molecules of } NADH.

Explanation:

One glucose molecule produces two pyruvate molecules, which lead to two turns of the Krebs cycle. Each turn of the cycle produces 2 CO22\text{ CO}_2 and 3 NADH3\text{ NADH}. Therefore, 2×2=4 CO22 \times 2 = 4\text{ CO}_2 and 2×3=6 NADH2 \times 3 = 6\text{ NADH}.