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Biology: Biochemistry and Plant Physiology - Enzymes as Biological Catalysts and Factors Affecting Enzyme Activity

Grade 8IB

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, which speed up chemical reactions in living organisms without being consumed in the process.

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The Lock and Key Hypothesis: Enzymes have a specific 3D shape with an active site where the substrate binds. Only a substrate with a complementary shape can fit, forming an enzyme-substrate complex.

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Activation Energy (EaE_a): Enzymes work by lowering the EaE_a required for a reaction to proceed, allowing metabolic processes to occur rapidly at body temperature.

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Effect of Temperature: As temperature increases, kinetic energy increases, leading to more frequent collisions between enzymes and substrates. However, beyond the optimum temperature (usually ≈37∘C\approx 37^\circ C for humans), the enzyme becomes denatured (the active site loses its shape).

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Effect of pH: Every enzyme has an optimum pH (e.g., Pepsin at pH≈2pH \approx 2, Amylase at pH≈7pH \approx 7). Deviations from this pH can break ionic and hydrogen bonds, leading to denaturation.

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Substrate Concentration: Increasing substrate concentration ([S][S]) increases the rate of reaction until all active sites are occupied (the saturation point, or VmaxV_{max}).

📐Formulae

Rate of Reaction=1tRate \ of \ Reaction = \frac{1}{t}

Rate=Δ[Product]ΔtRate = \frac{\Delta [Product]}{\Delta t}

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

💡Examples

Problem 1:

In an experiment investigating the breakdown of starch by the enzyme amylase, 10 cm310 \text{ cm}^3 of starch was completely digested in 5 minutes5 \text{ minutes} at 30∘C30^\circ C. Calculate the rate of reaction in cm3 min−1\text{cm}^3\text{ min}^{-1}.

Solution:

Rate=Volume of StarchTime=10 cm35 min=2 cm3 min−1Rate = \frac{\text{Volume of Starch}}{\text{Time}} = \frac{10 \text{ cm}^3}{5 \text{ min}} = 2 \text{ cm}^3\text{ min}^{-1}

Explanation:

The rate is determined by dividing the total amount of substrate processed by the time taken for the reaction to reach completion.

Problem 2:

If the rate of an enzyme-controlled reaction at 20∘C20^\circ C is 15 units/sec15 \text{ units/sec}, and the temperature coefficient Q10Q_{10} is 22, what would be the predicted rate at 30∘C30^\circ C?

Solution:

Rate30∘C=Rate20∘C×Q10=15×2=30 units/secRate_{30^\circ C} = Rate_{20^\circ C} \times Q_{10} = 15 \times 2 = 30 \text{ units/sec}

Explanation:

The Q10Q_{10} value represents the factor by which the reaction rate increases when the temperature is raised by 10∘C10^\circ C. Since the temperature increased from 20∘C20^\circ C to 30∘C30^\circ C, we multiply the initial rate by 22.