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Interaction and interdependence - Enzymes and metabolism

Grade 12IBBiology

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

🔑Concepts

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Metabolism is the totality of an organism's chemical reactions, consisting of catabolic pathways (breaking down molecules, e.g., cell respiration) and anabolic pathways (building molecules, e.g., photosynthesis).

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Enzymes are biological catalysts, typically proteins, that increase the rate of reaction by lowering the activation energy (EaE_a).

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The induced-fit model describes how the enzyme's active site undergoes a conformational change upon substrate binding to achieve a tighter fit, facilitating the conversion of substrates into products.

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Factors affecting enzyme activity include temperature, pH, and substrate concentration [S][S]. Extreme temperature and pH can cause denaturation, where the enzyme loses its tertiary structure and catalytic ability.

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Competitive inhibition occurs when a molecule similar to the substrate binds to the active site, which can be overcome by increasing [S][S].

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Non-competitive (allosteric) inhibition occurs when an inhibitor binds to a site other than the active site (allosteric site), changing the enzyme's shape so the substrate can no longer bind.

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Metabolic pathways are often regulated by end-product inhibition, where the final product of a chain acts as a non-competitive inhibitor for the first enzyme in the pathway.

📐Formulae

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

Q10=(R2R1)10T2−T1Q_{10} = \left( \frac{R_2}{R_1} \right)^{\frac{10}{T_2 - T_1}}

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

💡Examples

Problem 1:

In an experiment, an enzyme catalyzes the breakdown of 100 mmol100\text{ mmol} of substrate into products in 55 minutes. Calculate the initial rate of reaction in mmol min−1\text{mmol min}^{-1}.

Solution:

Rate=100 mmol5 min=20 mmol min−1\text{Rate} = \frac{100\text{ mmol}}{5\text{ min}} = 20\text{ mmol min}^{-1}

Explanation:

The rate of an enzyme-catalyzed reaction is determined by the change in concentration of substrate or product over a specific time interval.

Problem 2:

An enzyme-controlled reaction has a rate of 12 units/sec12\text{ units/sec} at 20∘C20^{\circ}C. If the Q10Q_{10} of the reaction is 22, what will be the rate at 30∘C30^{\circ}C?

Solution:

R2=R1×Q10T2−T110R_2 = R_1 \times Q_{10}^{\frac{T_2 - T_1}{10}} R2=12×230−2010=12×21=24 units/secR_2 = 12 \times 2^{\frac{30 - 20}{10}} = 12 \times 2^1 = 24\text{ units/sec}

Explanation:

The temperature coefficient Q10Q_{10} represents the factor by which the reaction rate increases with a 10∘C10^{\circ}C rise in temperature.