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Molecular Biology - Molecules to Metabolism

Grade 11IBBiology

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

🔑Concepts

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

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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.

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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}.

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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).

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

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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.

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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)}

R−OH+HO−R′→CondensationR−O−R′+H2OR-OH + HO-R' \xrightarrow{\text{Condensation}} R-O-R' + H_2O

R−O−R′+H2O→HydrolysisR−OH+HO−R′R-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=C12H24O12−H2O=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 C−NC-N peptide bond.