1 What is metabolism?

• The life of an organism depends on continuous biochemical reactions that take place in the presence of water inside its cells.
• All these reactions together are called metabolism.
• There are two kinds:

• A.
• Catabolic processes (catabolism). Chemical bonds between the atoms of complex molecules are broken to release the energy stored in them.
• Example: the breakdown of glucose in cellular respiration into carbon dioxide and water, releasing energy:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + energy (stored mainly as ATP, partly released as heat)

• B.
• Anabolic processes (anabolism). Large and complex molecules are built from simple molecules. These processes consume energy.
• Examples: photosynthesis and the synthesis of proteins from amino acids.

6 CO₂ + 6 H₂O + light energy → C₆H₁₂O₆ + 6 O₂ (in chloroplasts, with chlorophyll)

Amino acid + amino acid → dipeptide + H₂O (a water molecule is released each time a peptide bond forms; the reverse reaction, digestion of protein, uses water — hydrolysis).

• Water is a reactant in some reactions (photosynthesis, hydrolysis in digestion) and a product in others (respiration, condensation reactions that join amino acids, glucose units...).
• In every case the reactions occur in watery solution.

Problem. Calculate the mass of CO₂, water and oxygen involved when 90 g of glucose is completely oxidised in respiration. (C = 12, H = 1, O = 16; M(glucose) = 180.)

Solution. n(glucose) = 90/180 = 0.5 mol. From the equation 1 mol glucose needs 6 mol O₂ and gives 6 mol CO₂ and 6 mol H₂O.

• O₂ used = 3.0 mol × 32 = 96 g.
• CO₂ formed = 3.0 mol × 44 = 132 g.
• H₂O formed = 3.0 mol × 18 = 54 g. Check mass: 90 + 96 = 186 g = 132 + 54 ✔. The "metabolic water" produced (54 g from 90 g glucose) is a real water source for animals such as camels that rely on fat and carbohydrate oxidation in desert conditions.

• Problem. Plants in a lake take in 88 g of CO₂ during photosynthesis.
• Find the mass of glucose formed and of oxygen released, and the mass of water used.

Solution. n(CO₂) = 88/44 = 2.0 mol. 6 CO₂ → 1 C₆H₁₂O₆, so n(glucose) = 2/6 = 0.333 mol → mass = 0.333 × 180 = 60 g. O₂ released = 2.0 mol = 64 g. Water used = 2.0 mol = 36 g. Check: 88 + 36 = 124 g = 60 + 64 ✔.