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Enzymes - Enzyme action

Grade 11IGCSEBiology

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

🔑Concepts

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Enzymes are biological catalysts made of proteins that speed up the rate of metabolic reactions by lowering the activation energy (EaE_a).

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The Active Site is a specific region on the enzyme surface where the substrate (SS) binds. The shape of the active site is complementary to the substrate, known as the Lock and Key Hypothesis.

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When a substrate binds to the active site, it forms an temporary Enzyme-Substrate Complex (ESES).

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Temperature Influence: Increasing temperature increases the kinetic energy of molecules, leading to more frequent successful collisions. However, beyond the optimal temperature, enzymes undergo denaturation, where thermal energy breaks the bonds (like hydrogen bonds) maintaining the protein's 3D shape.

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pH Influence: Changes in pHpH affect the charges on the amino acids in the enzyme. Extreme pHpH levels (high [H+][H^+] or [OH−][OH^-] concentrations) can denature the enzyme, altering the active site and preventing the substrate from binding.

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Enzymes are chemically unchanged at the end of the reaction and can be reused: E+S→ES→E+PE + S \rightarrow ES \rightarrow E + P.

📐Formulae

E+S⇌ES→E+PE + S \rightleftharpoons ES \rightarrow E + P

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

Q10=Rate at (T+10)∘CRate at T∘CQ_{10} = \frac{\text{Rate at } (T + 10)^\circ C}{\text{Rate at } T^\circ C}

Rate∝1t\text{Rate} \propto \frac{1}{t}

💡Examples

Problem 1:

In an experiment using the enzyme Catalase, 20 cm320\text{ cm}^3 of O2O_2 gas was collected in 40 seconds40\text{ seconds}. Calculate the average rate of reaction.

Solution:

Rate=20 cm340 s=0.5 cm3/s\text{Rate} = \frac{20\text{ cm}^3}{40\text{ s}} = 0.5\text{ cm}^3/\text{s}

Explanation:

The rate is determined by dividing the total volume of product (O2O_2) by the time taken for the reaction to occur.

Problem 2:

An enzyme-controlled reaction has a rate of 12 units/s12\text{ units/s} at 25∘C25^\circ C. If the Q10Q_{10} for this enzyme is 2.02.0, what is the estimated rate at 35∘C35^\circ C?

Solution:

New Rate=12×2.0=24 units/s\text{New Rate} = 12 \times 2.0 = 24\text{ units/s}

Explanation:

The Q10Q_{10} (Temperature Coefficient) represents the factor by which the rate increases when the temperature is raised by 10∘C10^\circ C.

Problem 3:

Explain why the reaction rate of Salivary Amylase drops to zero when it reaches the stomach where the pHpH is approximately 2.02.0.

Solution:

The high concentration of H+H^+ ions at pH 2.0pH\ 2.0 disrupts the ionic and hydrogen bonds in the enzyme, leading to denaturation.

Explanation:

Amylase has an optimal pHpH of approximately 7.07.0. The acidic environment of the stomach changes the shape of the active site, so it is no longer complementary to the starch substrate.

Enzyme action - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology