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Biological Chemistry - Factors Affecting Enzyme Activity

Grade 9IB

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 by lowering the activation energy (EaE_a) required for the reaction to occur.

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Temperature: As temperature increases, the kinetic energy of molecules increases, leading to more frequent collisions between the enzyme and substrate. However, beyond the optimum temperature (ToptimumT_{optimum}), the enzyme's tertiary structure breaks down (denaturation), and the active site loses its shape.

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pH Level: Most enzymes have an optimum pH where they function most efficiently. Extreme acidic or alkaline conditions can disrupt the ionic and hydrogen bonds in the enzyme, leading to denaturation.

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Substrate Concentration: Increasing the concentration of substrate increases the rate of reaction up to a point. Once all active sites are occupied (saturation), the reaction reaches its maximum velocity (VmaxV_{max}), and further increases in substrate concentration will not affect the rate.

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Enzyme Concentration: If substrate is in excess, increasing the enzyme concentration linearly increases the rate of reaction because more active sites are available for catalysis.

📐Formulae

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

Rate≈1tRate \approx \frac{1}{t}

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

💡Examples

Problem 1:

In an experiment, 20 cm320\text{ cm}^3 of oxygen gas was produced by the breakdown of hydrogen peroxide using the enzyme catalase over a period of 40 seconds40\text{ seconds}. Calculate the average rate of reaction.

Solution:

Rate=Volume of O2Time TakenRate = \frac{\text{Volume of } O_2}{\text{Time Taken}} Rate=20 cm340 sRate = \frac{20\text{ cm}^3}{40\text{ s}} Rate=0.5 cm3/sRate = 0.5\text{ cm}^3/\text{s}

Explanation:

The rate is determined by dividing the total change in the amount of product (oxygen) by the time interval during which the reaction occurred.

Problem 2:

If an enzyme-controlled reaction has a rate of 12 units/min12\text{ units/min} at 20∘C20^\circ C and the Q10Q_{10} value is 2.02.0 for this range, what is the expected rate at 30∘C30^\circ C?

Solution:

Rate30∘C=Rate20∘C×Q10Rate_{30^\circ C} = Rate_{20^\circ C} \times Q_{10} Rate30∘C=12×2.0Rate_{30^\circ C} = 12 \times 2.0 Rate30∘C=24 units/minRate_{30^\circ C} = 24\text{ units/min}

Explanation:

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