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Molecular Biology - Enzymes

Grade 12IBBiology

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

🔑Concepts

Enzymes are globular proteins that act as biological catalysts by lowering the activation energy (ΔG\Delta G^{\ddagger} or EaE_a) of a chemical reaction without being consumed.

The active site is a specific region of the enzyme where the substrate binds. The 'Induced Fit Model' describes how the enzyme undergoes a conformational change to improve the fit with the substrate SS to form the enzyme-substrate complex (ESES).

Collision theory states that for a reaction to occur, the substrate must collide with the active site with sufficient energy and at the correct orientation. Factors like temperature (TT) and concentration affect these collision rates.

Factors affecting rate: (1) Temperature increases kinetic energy until the optimum; beyond this, thermal agitation breaks hydrogen bonds leading to denaturation. (2) pHpH affects the ionization of RR-groups, altering the protein's tertiary structure. (3) Substrate concentration [S][S] increases rate until all active sites are saturated (VmaxV_{max}).

Inhibition: Competitive inhibitors bind to the active site and can be overcome by increasing [S][S]. Non-competitive (allosteric) inhibitors bind to a site other than the active site, changing the enzyme's shape and reducing VmaxV_{max} regardless of [S][S].

Immobilized enzymes are often used in industry (e.g., using lactase to produce lactose-free milk). Advantages include easier separation of the product and increased stability against changes in TT and pHpH.

End-product inhibition is a form of negative feedback where the final product of a metabolic pathway acts as a non-competitive inhibitor for the first enzyme in the chain (e.g., Threonine to Isoleucine pathway).

📐Formulae

Rate of reaction=ΔProductΔt=ΔSubstrateΔt\text{Rate of reaction} = \frac{\Delta \text{Product}}{\Delta t} = -\frac{\Delta \text{Substrate}}{\Delta t}

Q10=(R2R1)10T2T1Q_{10} = \left( \frac{R_2}{R_1} \right)^{\frac{10}{T_2 - T_1}}

E+SESE+PE + S \rightleftharpoons ES \rightarrow E + P

V=Vmax[S]Km+[S]V = \frac{V_{max}[S]}{K_m + [S]}

💡Examples

Problem 1:

An experiment measuring the breakdown of hydrogen peroxide (H2O2H_2O_2) by catalase produced 20 cm320\text{ cm}^3 of oxygen (O2O_2) in 4040 seconds. Calculate the initial rate of reaction in cm3 s1\text{cm}^3\text{ s}^{-1}.

Solution:

Rate=20 cm340 s=0.5 cm3 s1\text{Rate} = \frac{20\text{ cm}^3}{40\text{ s}} = 0.5\text{ cm}^3\text{ s}^{-1}

Explanation:

The rate is determined by dividing the change in the volume of the product (O2O_2) by the time interval taken for the reaction.

Problem 2:

In a metabolic pathway where Threonine is converted to Isoleucine, how does the concentration of Isoleucine affect the enzyme Threonine Dehydratase?

Solution:

Isoleucine acts as a non-competitive inhibitor. As [Isoleucine][\text{Isoleucine}] increases, it binds to the allosteric site of Threonine Dehydratase, decreasing the reaction rate.

Explanation:

This is an example of end-product inhibition (negative feedback), ensuring that the cell does not overproduce Isoleucine when levels are already sufficient.

Enzymes - Revision Notes & Key Diagrams | IB Grade 12 Biology