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Molecular Biology - Molecules to Metabolism

Grade 11IBBiology

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

🔑Concepts

Molecular biology explains living processes in terms of the chemical substances involved. The most important elements are carbon, hydrogen, oxygen, and nitrogen.

Carbon atoms can form four covalent bonds, allowing a diversity of stable compounds to exist. This allows for the formation of complex structures like rings and chains.

Metabolism is the web of all enzyme-catalyzed reactions in a cell or organism. It is often summarized as: Metabolism=Anabolism+Catabolism\text{Metabolism} = \text{Anabolism} + \text{Catabolism}.

Anabolism is the synthesis of complex molecules from simpler molecules, including the formation of macromolecules from monomers by condensation reactions (e.g., GlycogenGlycogen from GlucoseGlucose).

Catabolism is the breakdown of complex molecules into simpler molecules, including the hydrolysis of macromolecules into monomers (e.g., GlucoseGlucose from StarchStarch).

Condensation reactions involve the removal of a water molecule (H2OH_2O) to form a covalent bond, while hydrolysis reactions require the addition of a water molecule (H2OH_2O) to break a covalent bond.

The falsification of vitalism occurred when Friedrich Wöhler synthesized urea (NH2CONH2NH_2CONH_2) artificially, proving that organic molecules do not require a 'vital force' to be created.

📐Formulae

Cn(H2O)n (General formula for monosaccharides)C_n(H_2O)_n \text{ (General formula for monosaccharides)}

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

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

NH2CONH2 (Urea)NH_2CONH_2 \text{ (Urea)}

ROH+HORCondensationROR+H2OR-OH + HO-R' \xrightarrow{\text{Condensation}} R-O-R' + H_2O

ROR+H2OHydrolysisROH+HORR-O-R' + H_2O \xrightarrow{\text{Hydrolysis}} R-OH + HO-R'

💡Examples

Problem 1:

Calculate the molecular formula of a disaccharide formed by the condensation of two hexose sugars with the formula C6H12O6C_6H_{12}O_6.

Solution:

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

Explanation:

When two glucose molecules (C6H12O6C_6H_{12}O_6) join, one molecule of water (H2OH_2O) is removed. Therefore: (2×C6H12O6)H2O=C12H24O12H2O=C12H22O11(2 \times C_6H_{12}O_6) - H_2O = C_{12}H_{24}O_{12} - H_2O = C_{12}H_{22}O_{11}.

Problem 2:

Identify the type of reaction that occurs when a triglyceride is broken down into glycerol and three fatty acids.

Solution:

Hydrolysis reaction.

Explanation:

Because a complex lipid is being broken down into simpler components, it is a catabolic process. Specifically, three molecules of H2OH_2O are consumed to break the ester bonds, making it a hydrolysis reaction.

Problem 3:

Which functional groups are involved in the formation of a peptide bond between two amino acids?

Solution:

The amine group (NH2-NH_2) and the carboxyl group (COOH-COOH).

Explanation:

In a condensation reaction, the OHOH from the carboxyl group of one amino acid and the HH from the amine group of another amino acid combine to form H2OH_2O, leaving a CNC-N peptide bond.

Molecules to Metabolism - Revision Notes & Key Diagrams | IB Grade 11 Biology