Biological Molecules

Two α-glucose molecules join to form maltose. What type of bond forms and what molecule is released?

• AA 1,6-glycosidic bond, releasing CO₂
• BAn ester bond, releasing H₂O
• CA 1,4-glycosidic bond, releasing H₂O✓
• DA peptide bond, releasing H₂O

Answer: Maltose forms when two α-glucose monomers join by a condensation reaction, creating a 1,4-glycosidic bond and releasing one molecule of water. Ester bonds occur in lipids and peptide bonds in proteins.

Cellulose and amylose are both polymers of glucose, yet have very different properties. Which statement explains the key structural difference?

• ACellulose is made of β-glucose with alternate monomers inverted, forming straight chains held by hydrogen bonds✓
• BCellulose is made of α-glucose forming a branched, coiled molecule
• CAmylose is made of β-glucose joined by 1,6-glycosidic bonds
• DCellulose contains nitrogen whereas amylose does not

Answer: Cellulose is a polymer of β-glucose; to form the 1,4 bond, alternate monomers are inverted, producing long straight chains that hydrogen-bond into strong microfibrils. Amylose is α-glucose, which forms a helical (coiled) chain. Neither contains nitrogen.

Which description correctly summarises the formation and bonding of a triglyceride?

• AOne glycerol and three fatty acids joined by three ester bonds, formed by condensation✓
• BOne glycerol and two fatty acids and a phosphate group joined by glycosidic bonds
• CThree glycerol molecules and one fatty acid joined by peptide bonds
• DOne fatty acid and three glycerol molecules joined by hydrogen bonds

Answer: A triglyceride forms when one glycerol molecule bonds with three fatty acids; each condensation reaction forms an ester bond and releases water, giving three ester bonds and three water molecules in total. A phospholipid has only two fatty acids plus a phosphate group.

Why are phospholipids able to form a bilayer in an aqueous environment?

• ABoth the heads and tails are hydrophobic
• BBoth the heads and tails are hydrophilic
• CThey have hydrophilic phosphate heads and hydrophobic fatty acid tails✓
• DThe fatty acid tails carry a negative charge that repels water

Answer: Phospholipids are amphipathic: the phosphate head is hydrophilic (polar) and the two fatty acid tails are hydrophobic. In water the heads face outward toward the aqueous solution and the tails point inward, forming a bilayer. The phosphate group, not the tails, carries the charge.

Which bond is responsible for maintaining the α-helix and β-pleated sheet arrangements that define the secondary structure of a protein?

• ADisulfide bridges
• BIonic bonds
• CPeptide bonds
• DHydrogen bonds✓

Answer: Secondary structure (α-helices and β-pleated sheets) is held together by hydrogen bonds between the C=O and N–H groups of the polypeptide backbone. Disulfide, ionic and hydrophobic interactions stabilise tertiary structure; peptide bonds form the primary sequence.

Which statement best describes the 'induced fit' model of enzyme action?

• AThe substrate is a perfect, rigid complement to the active site before binding
• BThe active site changes shape slightly as the substrate binds, fitting more closely around it✓
• CThe enzyme is permanently denatured each time it catalyses a reaction
• DThe substrate provides the activation energy for the reaction

Answer: In the induced fit model the active site is not an exact match initially; as the substrate binds, the active site moulds around it, straining bonds in the substrate and lowering the activation energy. This refines the older 'lock and key' model, which assumed a rigid, perfectly complementary active site.

A competitive inhibitor and a non-competitive inhibitor act on the same enzyme. Which statement is correct?

• AA competitive inhibitor binds the active site and its effect is reduced by increasing substrate concentration✓
• BA competitive inhibitor binds away from the active site and cannot be overcome by more substrate
• CA non-competitive inhibitor binds the active site, competing directly with the substrate
• DIncreasing substrate concentration reverses the effect of a non-competitive inhibitor

Answer: A competitive inhibitor has a similar shape to the substrate and binds the active site; raising substrate concentration outcompetes it, so its effect is reduced. A non-competitive inhibitor binds elsewhere (allosteric site), altering the active site, and its effect is NOT overcome by adding more substrate.

Many of water's biologically important properties arise from hydrogen bonding between molecules. Which property is NOT a direct consequence of water being a polar molecule that forms hydrogen bonds?

• AHigh specific heat capacity, stabilising temperatures
• BHigh latent heat of vaporisation, allowing cooling by evaporation
• CActing as a solvent for ionic and polar substances
• DBeing chemically inert and unable to take part in metabolic reactions✓

Answer: Water is highly reactive metabolically — it is a reactant in hydrolysis and photosynthesis and a product of condensation and respiration, so it is not inert. Its high heat capacity, high latent heat of vaporisation and solvent properties all result from polarity and hydrogen bonding.

Which inorganic ion is essential for the formation of haemoglobin and is found at the centre of each haem group?

• ASodium (Na⁺)
• BCalcium (Ca²⁺)
• CPhosphate (PO₄³⁻)
• DIron (Fe²⁺)✓

Answer: Each haem group contains a central Fe²⁺ ion to which oxygen binds reversibly, enabling haemoglobin to transport oxygen. Sodium is important in nerve impulses, calcium in bone and muscle contraction, and phosphate in nucleotides, ATP and phospholipids.

A student tests a solution for reducing sugar and then for a non-reducing sugar. The Benedict's test is negative, but after boiling the sample with dilute hydrochloric acid, neutralising with sodium hydrogencarbonate, and repeating Benedict's, a brick-red precipitate forms. What does this indicate?

• ALipid is present
• BA non-reducing sugar such as sucrose is present✓
• CStarch is present
• DProtein is present

Answer: A non-reducing sugar (e.g. sucrose) gives a negative Benedict's test until it is hydrolysed by boiling with acid into its reducing monosaccharides; after neutralising and retesting, a brick-red precipitate confirms a non-reducing sugar. Starch is detected with iodine, lipids by the emulsion test, and proteins by the biuret test.