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Enzymes - Enzyme action

Grade 12A LevelBiology

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

🔑Concepts

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Enzymes are globular proteins that act as biological catalysts, increasing the rate of metabolic reactions by lowering the activation energy (EaE_a).

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The 'Lock and Key' hypothesis states that the substrate shape is exactly complementary to the shape of the enzyme's active site.

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The 'Induced Fit' model suggests that the active site is flexible and undergoes a conformational change to fit the substrate more tightly upon binding, forming an enzyme-substrate complex (ESES complex).

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Factors affecting enzyme action include temperature, pHpH, substrate concentration ([S][S]), and enzyme concentration ([E][E]).

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Denaturation occurs when high temperatures or extreme pHpH levels break the hydrogen and ionic bonds holding the enzyme's tertiary structure, permanently altering the active site.

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Competitive inhibitors bind to the active site, while non-competitive inhibitors bind to an allosteric site, changing the enzyme's overall shape.

📐Formulae

Rate of Reaction=ΔProductΔt\text{Rate of Reaction} = \frac{\Delta \text{Product}}{\Delta t}

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

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

💡Examples

Problem 1:

A student measures the breakdown of hydrogen peroxide (H2O2H_2O_2) by the enzyme catalase. If 20 cm320\text{ cm}^3 of oxygen (O2O_2) is produced in 55 minutes, calculate the initial rate of reaction in cm3 min−1\text{cm}^3\text{ min}^{-1}.

Solution:

4 cm3 min−14\text{ cm}^3\text{ min}^{-1}

Explanation:

The rate of reaction is calculated by dividing the volume of product formed by the time taken: 20 cm35 min=4 cm3 min−1\frac{20\text{ cm}^3}{5\text{ min}} = 4\text{ cm}^3\text{ min}^{-1}.

Problem 2:

If the rate of an enzyme-catalyzed reaction is 1.5 mmol dm−3 s−11.5\text{ mmol dm}^{-3}\text{ s}^{-1} at 25∘C25^{\circ}C, what is the expected rate at 35∘C35^{\circ}C if the temperature coefficient (Q10Q_{10}) is 2.02.0?

Solution:

3.0 mmol dm−3 s−13.0\text{ mmol dm}^{-3}\text{ s}^{-1}

Explanation:

The Q10Q_{10} value indicates how much the rate increases with a 10∘C10^{\circ}C rise in temperature. Using the formula New Rate=Old Rate×Q10\text{New Rate} = \text{Old Rate} \times Q_{10}, we get 1.5×2.0=3.01.5 \times 2.0 = 3.0.