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Metabolism, Cell Respiration and Photosynthesis (AHL) - Metabolism

Grade 12IBBiology

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

🔑Concepts

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Metabolic pathways consist of chains and cycles of enzyme-catalyzed reactions. For example, glycolysis is a linear chain, while the Krebs cycle is a cyclical pathway.

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Enzymes catalyze chemical reactions by lowering the activation energy (EaE_a). This allows the reaction to proceed at a faster rate without increasing temperature.

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Competitive inhibitors are structurally similar to the substrate and bind to the active site. Their effect can be overcome by increasing substrate concentration [S][S], meaning VmaxV_{max} is reached but KmK_m increases.

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Non-competitive inhibitors bind to an allosteric site, causing a conformational change in the active site. This reduces the maximum rate of reaction (VmaxV_{max}), while the affinity for the substrate (KmK_m) remains unchanged.

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End-product inhibition is a form of negative feedback where the final product of a pathway acts as a non-competitive inhibitor for the first enzyme in the sequence. A classic example is the conversion of L−threonineL-threonine to L−isoleucineL-isoleucine.

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Bioinformatics involves using databases to locate and identify potential new anti-malarial drugs by targeting metabolic pathways in PlasmodiumPlasmodium falciparumfalciparum.

📐Formulae

Rate of reaction=ΔProduct ConcentrationΔt\text{Rate of reaction} = \frac{\Delta \text{Product Concentration}}{\Delta t}

v=Vmax[S]Km+[S]v = \frac{V_{max}[S]}{K_m + [S]}

ΔG=Gproducts−Greactants\Delta G = G_{\text{products}} - G_{\text{reactants}}

💡Examples

Problem 1:

Explain the effect of a competitive inhibitor on the reaction rate as substrate concentration increases compared to a control reaction.

Solution:

In the presence of a competitive inhibitor, the reaction rate increases more slowly at low [S][S] because the inhibitor competes for the active site. However, as [S]→∞[S] \to \infty, the probability of the substrate binding becomes much higher than the inhibitor, allowing the reaction to eventually reach the same VmaxV_{max} as the control. The value of KmK_m is higher in the inhibited reaction.

Explanation:

Competitive inhibition is reversible. Since the active site is not permanently occupied, high concentrations of substrate effectively 'out-compete' the inhibitor molecules.

Problem 2:

In the metabolic pathway converting Threonine to Isoleucine, identify the inhibitor and its mechanism.

Solution:

The inhibitor is L−isoleucineL-isoleucine and the mechanism is end-product (allosteric) inhibition.

Explanation:

When the concentration of L−isoleucineL-isoleucine builds up, it binds to the allosteric site of the enzyme threonine dehydratase. This changes the enzyme's conformation, preventing further conversion of threonine until isoleucine levels drop, thus maintaining homeostasis.

Problem 3:

Calculate the rate of reaction if the concentration of a product changes from 0.2 mol dm−30.2\text{ mol dm}^{-3} to 0.8 mol dm−30.8\text{ mol dm}^{-3} over a period of 120 seconds120\text{ seconds}.

Solution:

Rate=0.8−0.2120=0.6120=0.005 mol dm−3 s−1\text{Rate} = \frac{0.8 - 0.2}{120} = \frac{0.6}{120} = 0.005\text{ mol dm}^{-3}\text{ s}^{-1}

Explanation:

The rate is determined by the change in concentration (ΔProduct\Delta \text{Product}) divided by the time interval (Δt\Delta t).

Metabolism - Revision Notes & Key Diagrams | IB Grade 12 Biology