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Metabolism, Cell Respiration and Photosynthesis (AHL) - Metabolism

Grade 11IBBiology

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

🔑Concepts

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

Enzymes lower the activation energy (EaE_a) of the chemical reactions they catalyze, thereby increasing the reaction rate without being consumed.

Competitive inhibition involves a molecule other than the substrate binding to the enzyme's active site. The inhibitor is structurally similar to the substrate. Increasing substrate concentration can overcome this inhibition, eventually reaching the same maximum rate (VmaxV_{max}).

Non-competitive inhibition involves an inhibitor binding to an allosteric site (a site other than the active site). This changes the conformation of the enzyme, rendering the active site non-functional. The VmaxV_{max} is reduced regardless of substrate concentration.

End-product inhibition (a form of negative feedback) occurs when the final product of a metabolic pathway inhibits an enzyme that acts early in the pathway. An example is the conversion of LthreonineL-threonine to LisoleucineL-isoleucine.

Bioinformatics is used to identify potential new drug targets. Databases are searched for enzymes in pathogens (like the malaria parasite PlasmodiumPlasmodium falciparumfalciparum) that can be inhibited without affecting human metabolism.

📐Formulae

Rate of reaction=ΔProductΔTime\text{Rate of reaction} = \frac{\Delta \text{Product}}{\Delta \text{Time}}

Ea (with enzyme)<Ea (without enzyme)E_{a} \text{ (with enzyme)} < E_{a} \text{ (without enzyme)}

Total Enzyme=[E]+[ES]+[EI]\text{Total Enzyme} = [E] + [ES] + [EI]

💡Examples

Problem 1:

Contrast the effect of increasing substrate concentration [S][S] on the rate of reaction in the presence of a competitive inhibitor versus a non-competitive inhibitor.

Solution:

In competitive inhibition, increasing [S][S] allows the substrate to outcompete the inhibitor for the active site, meaning the VmaxV_{max} can still be reached. In non-competitive inhibition, the inhibitor reduces the total number of functional enzymes, so the VmaxV_{max} is significantly lowered and cannot be restored by adding more substrate.

Explanation:

Competitive inhibitors compete for the same site, so high substrate levels overwhelm the inhibitor. Non-competitive inhibitors disable the enzyme's function regardless of how much substrate is present.

Problem 2:

In the pathway: ThreonineEnzyme1AEnzyme2BEnzyme3CEnzyme4IsoleucineThreonine \xrightarrow{Enzyme 1} A \xrightarrow{Enzyme 2} B \xrightarrow{Enzyme 3} C \xrightarrow{Enzyme 4} Isoleucine. Explain what happens when IsoleucineIsoleucine levels are high.

Solution:

IsoleucineIsoleucine acts as a non-competitive inhibitor for Enzyme1Enzyme 1 (threonine deaminase). It binds to an allosteric site on Enzyme1Enzyme 1, changing its shape and preventing ThreonineThreonine from binding.

Explanation:

This is an example of end-product inhibition. It ensures that the cell does not waste energy and resources producing more IsoleucineIsoleucine than is currently required for protein synthesis.

Metabolism - Revision Notes & Key Diagrams | IB Grade 11 Biology