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Enzymes - Effects of temperature and pH

Grade 11IGCSEBiology

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

🔑Concepts

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Enzymes are biological catalysts made of protein molecules that speed up metabolic reactions by lowering the activation energy without being consumed in the process.

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The 'Lock and Key' hypothesis states that the substrate molecule has a complementary shape to the active site of the enzyme, forming an Enzyme-Substrate Complex (ESES Complex).

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Effect of Temperature: At low temperatures, molecules have low kinetic energy, resulting in fewer successful collisions. As temperature increases, the rate of reaction increases until it reaches the optimum temperature (usually ≈37∘C\approx 37^\circ C for human enzymes).

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Denaturation by Temperature: Beyond the optimum temperature, the increased thermal energy breaks the hydrogen bonds and other interactions holding the enzyme's tertiary structure. The active site changes shape, and the substrate can no longer fit.

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Effect of pH: Each enzyme has an optimum pHpH (e.g., Pepsin at pH≈2.0pH \approx 2.0, Trypsin at pH≈8.0pH \approx 8.0). Deviations from the optimum pHpH alter the ionization of amino acid R-groups, disrupting the ionic bonds that maintain the active site shape.

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Irreversibility: While low temperatures only slow down reactions, extreme pHpH or high temperatures usually cause irreversible denaturation of the protein structure.

📐Formulae

Rate of Reaction=1time (s)\text{Rate of Reaction} = \frac{1}{\text{time (s)}}

Rate=ΔProductΔTime\text{Rate} = \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}

Enzyme (E)+extSubstrate(S)⇌ES Complex→E+extProduct(P)\text{Enzyme (E)} + ext{Substrate (S)} \rightleftharpoons \text{ES Complex} \rightarrow \text{E} + ext{Product (P)}

💡Examples

Problem 1:

In an experiment, 10 cm310\text{ cm}^3 of starch is broken down by amylase. At 20∘C20^\circ C, the reaction takes 120 seconds120\text{ seconds}. At 30∘C30^\circ C, it takes 60 seconds60\text{ seconds}. Calculate the rate of reaction at 30∘C30^\circ C and determine the Q10Q_{10} value for this temperature range.

Solution:

  1. Rate at 30∘C=160≈0.0167 s−1\text{Rate at } 30^\circ C = \frac{1}{60} \approx 0.0167\text{ s}^{-1}.
  2. Rate at 20∘C=1120≈0.0083 s−1\text{Rate at } 20^\circ C = \frac{1}{120} \approx 0.0083\text{ s}^{-1}.
  3. Q10=0.01670.0083≈2Q_{10} = \frac{0.0167}{0.0083} \approx 2.

Explanation:

The rate is the reciprocal of the time taken for the substrate to disappear. The Q10Q_{10} coefficient of 22 indicates that the rate of reaction doubled with a 10∘C10^\circ C increase in temperature, which is typical for enzyme-controlled reactions below the optimum temperature.

Problem 2:

An enzyme-controlled reaction is measured at pH 2,pH 7,pH\ 2, pH\ 7, and pH 12pH\ 12. The reaction proceeds rapidly at pH 2pH\ 2 but does not occur at the other values. Identify the enzyme and explain the result.

Solution:

The enzyme is likely Pepsin. At pH 7pH\ 7 and pH 12pH\ 12, the concentration of H+H^+ ions is too low, which disrupts the ionic bonds in the enzyme's structure, causing denaturation.

Explanation:

Pepsin is a protease found in the stomach where conditions are highly acidic (HClHCl provides H+H^+ ions). Enzymes are sensitive to pHpH; if the pHpH shifts too far from the optimum, the change in charge on the amino acids causes the protein to unfold, destroying the active site.

Effects of temperature and pH - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology