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Biology - Enzymes (Properties and factors affecting activity)

Grade 9IGCSE

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

🔑Concepts

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Enzymes are biological catalysts that speed up chemical reactions in living organisms without being changed or consumed in the process. They are made of proteins.

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The Lock and Key Hypothesis: Enzymes have a specific 3D shape with an active site that is complementary to the shape of a specific substrate molecule. When they bind, they form an enzyme-substrate complex.

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Specificity: Because the active site has a unique shape, each enzyme can only catalyze one specific reaction or a group of very similar reactions.

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Effect of Temperature: As temperature increases, the kinetic energy of molecules increases, leading to more frequent successful collisions between the enzyme and substrate. However, beyond the optimum temperature (usually ≈37∘C\approx 37^\circ\text{C} in humans), the heat breaks the bonds holding the enzyme's shape, causing denaturation.

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Effect of pH: Every enzyme has an optimum pH (e.g., pH 22 for pepsin in the stomach, pH 77 for amylase). Moving too far from this pH alters the ionic charges of the amino acids, causing the enzyme to denature and the active site to lose its shape.

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Denaturation: This is a permanent change in the shape of the active site. Once denatured, the substrate can no longer fit, and the enzyme ceases to function. It is not 'killed' because enzymes are not alive.

📐Formulae

Rate of reaction=Change in amount of substrate or productTime taken\text{Rate of reaction} = \frac{\text{Change in amount of substrate or product}}{\text{Time taken}}

Rate∝1Time taken\text{Rate} \propto \frac{1}{\text{Time taken}}

Q10=Rate at (T+10)∘CRate at T∘C≈2 (for many biological reactions below optimum)Q_{10} = \frac{\text{Rate at } (T + 10)^\circ\text{C}}{\text{Rate at } T^\circ\text{C}} \approx 2 \text{ (for many biological reactions below optimum)}

💡Examples

Problem 1:

In an experiment investigating the breakdown of hydrogen peroxide (H2O2H_2O_2) by the enzyme catalase, 20 cm320\text{ cm}^3 of oxygen gas was collected in 40 seconds40\text{ seconds}. Calculate the average rate of reaction.

Solution:

Rate=20 cm340 s=0.5 cm3s−1\text{Rate} = \frac{20\text{ cm}^3}{40\text{ s}} = 0.5\text{ cm}^3\text{s}^{-1}

Explanation:

The rate is calculated by dividing the total volume of product formed by the time taken. The unit is cm3/s\text{cm}^3\text{/s} or cm3s−1\text{cm}^3\text{s}^{-1}.

Problem 2:

An enzyme-controlled reaction takes 10 minutes10\text{ minutes} at 20∘C20^\circ\text{C}. When the temperature is increased to 30∘C30^\circ\text{C}, the reaction takes only 5 minutes5\text{ minutes}. Explain this result using the kinetic theory.

Solution:

Rate at 20∘C=110=0.1 min−1\text{Rate at } 20^\circ\text{C} = \frac{1}{10} = 0.1\text{ min}^{-1} Rate at 30∘C=15=0.2 min−1\text{Rate at } 30^\circ\text{C} = \frac{1}{5} = 0.2\text{ min}^{-1}

Explanation:

The rate doubled because the temperature increase of 10∘C10^\circ\text{C} provided more kinetic energy to the molecules. This resulted in faster movement and more frequent successful collisions between the enzyme's active site and the substrate.