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Biological Chemistry - Enzymes as Biological Catalysts

Grade 9IB

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

🔑Concepts

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Enzymes are biological catalysts, which are primarily proteins made of long chains of amino acids folded into specific three-dimensional shapes.

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A catalyst is a substance that increases the rate of a chemical reaction without being permanently changed or consumed in the process.

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Enzymes work by lowering the activation energy (EaE_a), which is the minimum energy required for a reaction to occur.

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The specific region of the enzyme where the substrate binds is called the active site. Its shape is complementary to the substrate, often explained by the 'Lock and Key' hypothesis or the 'Induced Fit' model.

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The interaction between the enzyme (EE) and substrate (SS) results in the formation of an unstable intermediate known as the Enzyme-Substrate Complex (ESCESC).

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Factors affecting enzyme activity include temperature, pH, and substrate concentration. Most human enzymes have an optimum temperature (ToptT_{opt}) of approximately 37∘C37^\circ C.

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Denaturation occurs when high temperatures or extreme pH levels break the bonds (like hydrogen bonds) holding the enzyme's shape, causing the active site to deform so it can no longer bind the substrate.

📐Formulae

Rate of Reaction=Change in Concentration of Product or SubstrateTime taken\text{Rate of Reaction} = \frac{\text{Change in Concentration of Product or Substrate}}{\text{Time taken}}

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

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

💡Examples

Problem 1:

During an experiment, the enzyme amylase breaks down starch into maltose. If 0.5 g0.5 \text{ g} of starch is digested in 200 seconds200 \text{ seconds}, calculate the average rate of the reaction in mg/s\text{mg/s}.

Solution:

First, convert the mass from grams to milligrams: 0.5 g=0.5×1000=500 mg0.5 \text{ g} = 0.5 \times 1000 = 500 \text{ mg}. Now, use the rate formula: Rate=500 mg200 s\text{Rate} = \frac{500 \text{ mg}}{200 \text{ s}} Rate=2.5 mg/s\text{Rate} = 2.5 \text{ mg/s}

Explanation:

The rate of reaction measures how much substrate is converted per unit of time. In this case, we divided the total mass of the substrate processed by the total time in seconds.

Problem 2:

A student measures the rate of an enzyme-controlled reaction at 20∘C20^\circ C to be 15 units/min15 \text{ units/min} and at 30∘C30^\circ C to be 30 units/min30 \text{ units/min}. Calculate the temperature coefficient (Q10Q_{10}) for this range.

Solution:

Using the formula for Q10Q_{10}: Q10=Rate at 30∘CRate at 20∘CQ_{10} = \frac{\text{Rate at } 30^\circ C}{\text{Rate at } 20^\circ C} Q10=3015Q_{10} = \frac{30}{15} Q10=2Q_{10} = 2

Explanation:

The Q10Q_{10} value represents the factor by which the reaction rate increases when the temperature is raised by 10∘C10^\circ C. A value of 22 indicates the rate doubled.