explain the relative energy values of carbohydrates, lipids and proteins as respiratory substrates

Published by Patrick Mutisya · 14 days ago

Cambridge A‑Level Biology – Energy: Respiratory Substrates

Energy – Respiratory Substrates

Learning Objective

Explain the relative energy values of carbohydrates, lipids and proteins when they are used as respiratory substrates.

Key Concepts

  • Oxidation of organic molecules releases free energy that is captured as ATP.

  • Different macromolecules contain different amounts of reduced carbon, influencing the total ATP that can be generated.

  • Energy yield can be expressed per mole of substrate, per gram of substrate, or in kilojoules/kilocalories.

  • The efficiency of ATP synthesis is limited by the P/O ratio (approximately 2.5 ATP per NADH and 1.5 ATP per FADH₂).

Energy Yield per Molecule of Substrate

Substrate (example)Typical oxidation pathwayATP produced per moleEnergy (kJ mol⁻¹)Energy per gram (kJ g⁻¹)ATP equivalents per gram
Carbohydrate – glucose (C₆H₁₂O₆)Glycolysis → Pyruvate dehydrogenase → TCA cycle → Oxidative phosphorylation≈ 30–32 ATP≈ 880 kJ mol⁻¹≈ 17 kJ g⁻¹≈ 0.56 mol ATP g⁻¹
Lipid – palmitic acid (C₁₆H₃₂O₂)β‑oxidation → Acetyl‑CoA → TCA cycle → Oxidative phosphorylation≈ 106 ATP≈ 2 400 kJ mol⁻¹≈ 37 kJ g⁻¹≈ 1.21 mol ATP g⁻¹
Protein – average amino acid (e.g., alanine)Deamination → Entry as pyruvate, acetyl‑CoA or TCA intermediates → Oxidative phosphorylation≈ 20–25 ATP≈ 600 kJ mol⁻¹≈ 17 kJ g⁻¹≈ 0.56 mol ATP g⁻¹

Why Lipids Yield More Energy

Lipids contain long chains of reduced carbon atoms. Each round of β‑oxidation removes a two‑carbon acetyl‑CoA unit and generates one NADH and one FADH₂, both of which feed electrons into the electron‑transport chain. Consequently, a single fatty‑acid molecule can produce many more reducing equivalents than a carbohydrate of comparable mass.

Conversion Between Energy Units

The free‑energy change for hydrolysis of one mole of ATP under cellular conditions is about \$ΔG_{ATP} ≈ -30.5\ \text{kJ mol}^{-1}\$ (≈ 7.3 kcal mol⁻¹). Using this value, the ATP equivalents per gram shown in the table are calculated as:

\$\text{ATP per gram} = \frac{\text{Energy per gram (kJ g⁻¹)}}{ΔG_{ATP}}\$

Practical Implications for Metabolism

  1. During prolonged, low‑intensity exercise, the body preferentially oxidises lipids because they provide a large amount of ATP per gram.

  2. During high‑intensity, short‑duration activity, carbohydrates are favoured because they can be metabolised more rapidly, despite a lower ATP yield per gram.

  3. Proteins are generally a last‑resort substrate; they must first be deaminated, and their catabolism also supplies nitrogen for excretion.

Suggested diagram: Flow of carbon from carbohydrate, lipid and protein catabolism into the citric‑acid cycle and oxidative phosphorylation.