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9700 · 2.2

Carbohydrates and lipids — practice questions

Practice and worked examples for 9700 Carbohydrates and lipids. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Describe how the structural properties of cellulose contribute to its role in providing support to plant cell walls.

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  1. Monomer and Linkage: Cellulose is a polymer of β-glucose monomers, linked by β-1,4 glycosidic bonds. This specific linkage causes adjacent glucose units to be rotated 180° relative to each other.
  2. Straight Chains: This rotation prevents coiling and results in long, straight, unbranched chains of cellulose molecules.
  3. Hydrogen Bonding: These parallel chains can lie close together, allowing numerous weak hydrogen bonds to form between the hydroxyl (-OH) groups on adjacent chains.
  4. Microfibrils: The cumulative effect of thousands of hydrogen bonds bundles hundreds of these cellulose chains together to form strong, rope-like structures called microfibrils.
  5. High Tensile Strength: These microfibrils have very high tensile strength, resisting stretching forces. They are arranged in layers within the cell wall, providing rigidity and preventing the cell from bursting due to osmotic pressure (turgor).

Worked example 2

A triglyceride is formed from one molecule of glycerol (C₃H₈O₃) and three molecules of palmitic acid (C₁₆H₃₂O₂). Calculate the molecular mass of this triglyceride. (Relative atomic masses: C=12.0, H=1.0, O=16.0).

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  1. Calculate the molecular mass of the reactants:

    • Glycerol (C₃H₈O₃): (3 × 12.0) + (8 × 1.0) + (3 × 16.0) = 36.0 + 8.0 + 48.0 = 92.0
    • Palmitic Acid (C₁₆H₃₂O₂): (16 × 12.0) + (32 × 1.0) + (2 × 16.0) = 192.0 + 32.0 + 32.0 = 256.0
    • Total mass of 3 palmitic acid molecules: 3 × 256.0 = 768.0
    • Total mass of all reactants: 92.0 (glycerol) + 768.0 (3 fatty acids) = 860.0

  2. Account for water molecules lost:

    • The formation of a triglyceride involves three condensation reactions, forming three ester bonds.
    • Each condensation reaction releases one molecule of water (H₂O).
    • Therefore, 3 molecules of water are lost.

  3. Calculate the mass of water lost:

    • Molecular mass of H₂O: (2 × 1.0) + 16.0 = 18.0
    • Total mass of 3 water molecules: 3 × 18.0 = 54.0

  4. Calculate the final molecular mass of the triglyceride:

    • Subtract the mass of the water lost from the total mass of the reactants.
    • Final Molecular Mass: 860.0 - 54.0 = 806.0

    The molecular mass of the triglyceride is 806.0.

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