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Molecular Biology - Enzymes

Grade 11IBBiology

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 reaction by lowering the activation energy (EaE_a) without being consumed in the process.

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The active site is a specific region on the enzyme where the substrate binds; its specificity is determined by the tertiary structure of the protein.

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The 'Induced Fit Model' suggests that the enzyme's active site undergoes a conformational change upon substrate binding to achieve a tighter fit, rather than being a rigid 'Lock and Key'.

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Enzyme catalysis involves molecular motion and the collision of substrates with the active site. The rate of reaction is affected by the frequency of effective collisions.

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Denaturation is a structural change in the enzyme (usually caused by extreme pHpH or high temperature) that results in the loss of its biological properties and the distortion of the active site.

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Factors affecting enzyme activity include temperature, pHpH, and substrate concentration ([S][S]). As [S][S] increases, the rate increases until the active sites become saturated (VmaxV_{max}).

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Immobilized enzymes are used in industry, such as the use of lactase to produce lactose-free milk by breaking down lactose into glucose and galactose: C12H22O11+H2O→LactaseC6H12O6+C6H12O6C_{12}H_{22}O_{11} + H_2O \xrightarrow{Lactase} C_6H_{12}O_6 + C_6H_{12}O_6.

📐Formulae

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

Rate of Reaction=−Δ[Substrate]Δt\text{Rate of Reaction} = \frac{-\Delta [\text{Substrate}]}{\Delta t}

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

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

💡Examples

Problem 1:

Calculate the rate of an enzyme-catalyzed reaction if the concentration of the product increases from 0 mol dm−30 \text{ mol dm}^{-3} to 0.15 mol dm−30.15 \text{ mol dm}^{-3} over a period of 300300 seconds.

Solution:

Rate=0.15 mol dm−3−0 mol dm−3300 s=0.0005 mol dm−3 s−1\text{Rate} = \frac{0.15 \text{ mol dm}^{-3} - 0 \text{ mol dm}^{-3}}{300 \text{ s}} = 0.0005 \text{ mol dm}^{-3} \text{ s}^{-1} (or 5×10−4 mol dm−3 s−15 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1})

Explanation:

The rate of reaction is defined as the change in concentration of a product per unit time. We divide the total change in product concentration by the time interval.

Problem 2:

Explain why the rate of reaction levels off at high substrate concentrations [S][S].

Solution:

At high [S][S], the rate reaches VmaxV_{max}.

Explanation:

As the substrate concentration increases, more active sites are occupied. Eventually, every enzyme molecule's active site is working at its maximum capacity (saturated). At this point, adding more substrate cannot increase the rate further because the enzyme concentration becomes the limiting factor.

Enzymes - Revision Notes & Key Diagrams | IB Grade 11 Biology