krit.club logo

Biology - Enzymes

Grade 10IGCSE

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

🔑Concepts

•

Enzymes are biological catalysts made of proteins that speed up metabolic reactions by lowering the activation energy (EaE_a).

•

The 'Lock and Key' hypothesis states that the substrate shape is complementary to the enzyme's active site, forming an enzyme-substrate complex (ESES).

•

Temperature affects enzyme activity: as temperature increases, kinetic energy increases, leading to more frequent successful collisions until the optimum temperature (ToptT_{opt}) is reached.

•

Denaturation occurs when high temperatures or extreme pHpH values break the bonds holding the enzyme's protein structure, permanently changing the shape of the active site.

•

Enzymes are specific; for example, Amylase breaks down starch into maltose, Protease breaks down proteins into amino acids, and Lipase breaks down lipids into fatty acids and glycerol (C3H8O3C_3H_8O_3).

•

The rate of reaction is influenced by substrate concentration until the enzymes become saturated, reaching VmaxV_{max}.

📐Formulae

Rate of Reaction=Change in mass or volumeTime taken\text{Rate of Reaction} = \frac{\text{Change in mass or volume}}{\text{Time taken}}

Rate=1t\text{Rate} = \frac{1}{t}

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

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

💡Examples

Problem 1:

In an investigation into the effect of temperature on amylase, it was found that 10 cm310\text{ cm}^3 of starch was completely digested in 200 s200\text{ s} at 30∘C30^\circ\text{C} and in 100 s100\text{ s} at 40∘C40^\circ\text{C}. Calculate the rate of reaction at 40∘C40^\circ\text{C}.

Solution:

Rate=Volume of StarchTime=10 cm3100 s=0.1 cm3s−1\text{Rate} = \frac{\text{Volume of Starch}}{\text{Time}} = \frac{10\text{ cm}^3}{100\text{ s}} = 0.1\text{ cm}^3\text{s}^{-1}

Explanation:

The rate of reaction is calculated by dividing the amount of substrate acted upon by the time taken. At 40∘C40^\circ\text{C}, the reaction is twice as fast as at 30∘C30^\circ\text{C} due to increased kinetic energy.

Problem 2:

Explain why the rate of reaction for an enzyme drops to zero when the solution is heated to 80∘C80^\circ\text{C}.

Solution:

At 80∘C80^\circ\text{C}, the thermal energy is high enough to break the hydrogen and ionic bonds maintaining the tertiary structure of the protein. This causes denaturation.

Explanation:

When an enzyme denatures, the active site loses its specific shape. The substrate can no longer fit into the active site (it is no longer complementary), so no ESES complexes can form.