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Molecular Biology - Molecules to metabolism

Grade 12IBBiology

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

🔑Concepts

Molecular biology explains living processes in terms of the chemical substances involved, focusing on the structures and interactions of molecules.

Carbon atoms can form four covalent bonds, allowing for a vast diversity of stable compounds such as carbohydrates, lipids, proteins, and nucleic acids.

Metabolism is the totality of an organism's chemical reactions, categorized into anabolism and catabolism.

Anabolism is the synthesis of complex molecules from simpler ones, typically requiring energy (ATPATP). It involves condensation reactions where a water molecule (H2OH_2O) is released.

Catabolism is the breakdown of complex molecules into simpler ones, releasing energy. It involves hydrolysis reactions where a water molecule (H2OH_2O) is consumed to break bonds.

The theory of Vitalism stated that organic molecules could only be synthesized by living systems. This was falsified by Friedrich Wöhler in 1828 when he synthesized urea ((NH2)2CO(NH_2)_2CO) from inorganic ammonium cyanate (NH4CNONH_4CNO).

Carbohydrates are composed of carbon, hydrogen, and oxygen, usually in a ratio of 1:2:11:2:1. Common examples include glucose (C6H12O6C_6H_{12}O_6) and ribose (C5H10O5C_5H_{10}O_5).

📐Formulae

C6H12O6 (Glucose)C_6H_{12}O_6 \text{ (Glucose)}

C5H10O5 (Ribose)C_5H_{10}O_5 \text{ (Ribose)}

Monomer+MonomerCondensationDimer+H2O\text{Monomer} + \text{Monomer} \xrightarrow{\text{Condensation}} \text{Dimer} + H_2O

Polymer+nH2OHydrolysis(n+1)Monomers\text{Polymer} + nH_2O \xrightarrow{\text{Hydrolysis}} (n+1)\text{Monomers}

(NH2)2CO (Urea)(NH_2)_2CO \text{ (Urea)}

💡Examples

Problem 1:

Calculate the molecular formula of a disaccharide formed by the condensation of two glucose molecules (C6H12O6C_6H_{12}O_6).

Solution:

C12H22O11C_{12}H_{22}O_{11}

Explanation:

When two glucose molecules (2×C6H12O6=C12H24O122 \times C_6H_{12}O_6 = C_{12}H_{24}O_{12}) undergo a condensation reaction to form maltose, one molecule of H2OH_2O is removed. Therefore, the calculation is C12H24O12H2O=C12H22O11C_{12}H_{24}O_{12} - H_2O = C_{12}H_{22}O_{11}.

Problem 2:

Identify whether the following reaction is anabolic or catabolic: Triglyceride+3H2OGlycerol+3Fatty Acids\text{Triglyceride} + 3H_2O \rightarrow \text{Glycerol} + 3 \text{Fatty Acids}.

Solution:

Catabolic (Hydrolysis)

Explanation:

The reaction involves the addition of water (H2OH_2O) to break down a complex lipid (triglyceride) into its smaller components (glycerol and fatty acids). This process of breaking down macromolecules is defined as catabolism.

Molecules to metabolism - Revision Notes & Key Diagrams | IB Grade 12 Biology