Biological Molecules

Describe the bonding in water molecules and explain why ice is less dense than liquid water.

• Water is polar (O is S⁻, H is S⁺). Hydrogen bonds form between molecules. In ice, H-bonds hold molecules further apart in a rigid lattice structure, making ice less dense than liquid water (insulating aquatic habitats).

Explain the biological importance of the High Specific Heat Capacity and High Latent Heat of Vaporisation of water.

• High Specific Heat Capacity: A lot of energy is required to raise the temperature. This provides a thermally stable environment for aquatic organisms and allows enzymes to function efficiently.

• High Latent Heat of Vaporisation: A lot of energy is required to evaporate water. This provides a potent cooling effect (e.g., sweating/transpiration) with minimal fluid loss.

Draw/Describe the difference between a-glucose and B-glucose.

They are structural isomers.

• alpha-glucose: The hydroxyl (-OH) group on Carbon 1 is below the ring.

• beta-glucose: The hydroxyl (-OH) group on Carbon 1 is above the ring.

Name the monomers and the bond formed for: Maltose, Sucrose, and Lactose.

• Maltose: alpha-glucose + alpha-glucose.

• Sucrose: Glucose + Fructose.

• Lactose: Glucose + Galactose.

• Bond: All formed by condensation reactions creating glycosidic bonds.

Compare the structure of Amylose and Amylopectin.

• Amylose: Unbranched, 1-4 glycosidic bonds only. Coils into a helix (compact for storage).

• Amylopectin: Branched, 1-4 AND 1-6 glycosidic bonds. Branches allow many ends for rapid hydrolysis by enzymes.

How does the structure of Glycogen relate to its function in animals compared to Starch?

• Glycogen is more branched than amylopectin (more 1-6 bonds). This makes it more compact and provides more free ends for very rapid hydrolysis to release glucose for the high metabolic rate of animals.

Explain how the structure of cellulose provides high tensile strength.

• Polymer of beta-glucose.

• Alternate molecules are inverted (180°) to allow 1-4 glycosidic bond formation.

• Forms straight, unbranched chains.

• Chains run parallel and are cross-linked by many hydrogen bonds to form microfibrils.

Describe the formation of a triglyceride.

• One molecule of glycerol joins with three fatty acids via condensation reactions. Three ester bonds are formed, and three water molecules are released.

State two structural differences between triglycerides and phospholipids and explain how this affects phospholipid function.

• Difference 1: Phospholipids have 2 fatty acids, Triglycerides have 3.

• Difference 2: Phospholipids have a phosphate group replacing the third fatty acid.

• Function: The phosphate head is hydrophilic and fatty acid tails are hydrophobic (amphipathic), allowing them to form a bilayer in membranes (barrier to water-soluble substances).

What is the structural difference between saturated and unsaturated fatty acids?

• Saturated: No double bonds (C=C) between carbon atoms in the hydrocarbon chain.

• Unsaturated: Contain one (mono) or more (poly) C=C double bonds, causing a "kink" in the chain (lowering melting point).

How do Phospholipids and Cholesterol contribute to membrane structure?

• Phospholipids: Have a hydrophilic phosphate head and hydrophobic fatty acid tails. They form a bilayer (barrier to water-soluble substances).

• Cholesterol: Small, hydrophobic molecule that sits between tails. Regulates membrane fluidity (prevents freezing/too much movement) and stability.

What are the three functional groups attached to the central carbon of an amino acid?

• Amine group (NH₂).

• Carboxyl group (COOH)

• R-group (variable side chain that determines properties).

Define Primary, Secondary, Tertiary, and Quaternary structure.

• Primary: Sequence of amino acids (peptide bonds).

• Secondary: The coiling into an alpha-helix or folding into a beta-pleated sheet due to hydrogen bonds between the -NH and -CO groups of the peptide backbone.

• Tertiary: The final 3D shape formed by interactions between R-groups (Hydrophobic/hydrophilic interactions, Hydrogen bonds, Ionic bonds, Disulfide bonds).

Compare the properties and functions of Collagen and Haemoglobin.

• Collagen (Fibrous): Insoluble; structural (mechanical strength); composed of 3 polypeptide chains wound in a triple helix; cross-linked by covalent bonds.

• Haemoglobin (Globular): Soluble; metabolic (transport); specific spherical 3D shape; conjugated (contains prosthetic haem group Fe²⁺.

Give one biological role for each of the following ions: Ca²⁺, Na⁺, H⁺, Cl^-

• Ca²⁺: Nerve impulse transmission / Muscle contraction / Bone hardness / Cofactor for blood clotting.

• Na⁺: Co-transport of glucose/amino acids / Generating nerve impulses (action potentials).

• H⁺: Determining pH / Chemiosmosis (ATP synthesis) / Oxygen dissociation (Bohr effect).

• Cl^-: Cofactor for amylase / Chloride shift in blood (gas transport).

Describe the tests for: Starch, Reducing Sugar, Protein, Lipid.

• Starch: Iodine (Yellow/Brown → Blue/Black).

• Reducing Sugar: Benedict's + Heat (Blue → Green/Yellow/Orange/Brick Red).

• Protein: Biuret (Blue → Lilac/Purple).

Lipid: Emulsion test (Ethanol + Water → White Emulsion) .

How do you test for a non-reducing sugar (e.g., sucrose)?

1. Confirm negative result with Benedict's.

2. Boil new sample with dilute HCl (hydrolyses bonds).

3. Neutralise with Sodium Hydrogencarbonate (NaHCO₃).

4. Retest with Benedict's + Heat (Positive result = Brick Red).

How do you calculate an R_f value and what is it used for?

R_f = \frac{\text{Distance moved by solute}}{\text{Distance moved by solvent front}}

• Use: To identify separated biological molecules (e.g., amino acids) by comparing calculated values to known standards.

How can you determine the concentration of a reducing sugar solution quantitatively?

1. Perform Benedict's test (standardised volume/time).

2. Filter to remove precipitate.

3. Use a Colorimeter with a red filter to measure % transmission of the remaining filtrate (more sugar = less blue = higher transmission).

4. Compare to a calibration curve made from known concentrations.

Glycogen and cellulose are both carbohydrates. Describe two differences between the structure of a cellulose molecule and a glycogen molecule.(5 marks)

• Cellulose is made of beta-glucose, whereas glycogen is made of alpha-glucose.

• Cellulose has 1-4 glycosidic bonds only (straight chain), whereas glycogen has 1-4 and 1-6 glycosidic bonds (branched).

• Cellulose has straight chains, whereas glycogen is coiled/branched.

• Cellulose has alternate molecules inverted, glycogen does not.

Describe how the tertiary structure of a protein is held in place.(4 marks)

• It is held by interactions between R-groups (side chains).

• Disulfide bonds: Strong covalent bonds between sulfur atoms (cysteine).

• Ionic bonds: Between oppositely charged R-groups.

• Hydrogen bonds: Weak bonds between polar R-groups.

• Hydrophobic/Hydrophilic interactions: Hydrophobic R-groups cluster in the centre.

Water is a good solvent. Explain why water is a good solvent and how this aids the survival of organisms.(3 marks)

• Water is polar / has a dipole.

• It attracts/binds to charged/polar molecules (e.g., ions, glucose).

• Survival: Allows the transport of dissolved substances (e.g., in blood/sap/xylem) OR allows metabolic reactions to occur in solution.

Explain why triglycerides are excellent energy storage molecules.(3 marks)

• They have a high ratio of C-H bonds to carbon atoms (energy rich).

• They are insoluble in water, so they do not affect the water potential (Psi / osmosis of cells.

• They release more energy per gram than carbohydrates.

student separates amino acids using thin-layer chromatography. Why must the origin line be drawn in pencil, and why must the solvent level be below this line?(4 marks)

• Pencil: Ink would dissolve in the solvent / separate / run / mix with the amino acids (confusing results). Pencil is insoluble.

• Solvent level: If the solvent is above the line, the amino acids would dissolve directly into the solvent (in the beaker) rather than moving up the plate.