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 held by 1,6-glycosidic bonds
• CAmylose is made of β-glucose monomers joined end to end 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 three peptide bonds
• DThree fatty acids bonded directly to one another, with glycerol released as water

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 phosphate heads and the fatty acid tails are hydrophobic, repelling water
• BBoth the phosphate heads and the fatty acid 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, so it can be used just once
• DThe substrate provides the activation energy needed 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 to an allosteric site away from the active site, so its effect cannot be overcome by adding more substrate
• CA non-competitive inhibitor binds to the active site itself, competing directly with the substrate for the same position
• DIncreasing substrate concentration reverses the effect of a non-competitive inhibitor by displacing it from the enzyme

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.

Which bond forms between two monosaccharides during condensation?

• AA peptide bond, releasing one water molecule
• BA glycosidic bond, releasing one water molecule✓
• CAn ester bond, releasing one water molecule
• DA hydrogen bond, releasing one water molecule

Answer: All four are condensation reactions releasing water, so the water is not the clue — the bond name is. Sugars join by glycosidic bonds; peptide bonds join amino acids and ester bonds join glycerol to fatty acids.

Which test gives a brick-red precipitate with a reducing sugar?

• ABenedict's test, heated in a water bath✓
• BBiuret test, carried out at room temperature
• CEmulsion test, using ethanol then water
• DIodine test, using potassium iodide solution

Answer: Benedict's reagent is reduced on heating and shifts blue → green → yellow → orange → brick-red as sugar concentration rises. Biuret detects protein, the emulsion test lipid and iodine starch.

Why are unsaturated fatty acids liquid at room temperature?

• ADouble bonds shorten the chains so they pack loosely
• BDouble bonds kink the chains so they pack loosely✓
• CSingle bonds kink the chains so they pack tightly
• DSingle bonds lengthen the chains so they pack tightly

Answer: A C=C double bond puts a permanent kink in the hydrocarbon tail, so molecules cannot lie close together and intermolecular forces are weaker. Saturated chains are straight, pack tightly and are solid at room temperature.

What makes cellulose suited to a structural role in plant cell walls?

• AStraight chains hydrogen-bonded into strong microfibrils✓
• BBranched chains hydrogen-bonded into strong microfibrils
• CStraight chains held together by weak ionic attractions
• DCoiled chains cross-linked by strong covalent bridges

Answer: Alternate glucose units are inverted, so the chains are straight and lie parallel, cross-linking by many hydrogen bonds into microfibrils of great tensile strength. Branching and coiling describe glycogen and starch, which store energy rather than bear load.

Why does heating past the optimum reduce the rate of an enzyme-catalysed reaction?

• ABonds holding the primary structure break, altering the active site
• BSubstrate molecules gain too much energy to bind to the active site
• CBonds holding the tertiary structure break, altering the active site✓
• DThe enzyme is consumed more rapidly than the cell can replace it

Answer: Heat disrupts the hydrogen and ionic bonds maintaining the tertiary structure, so the active site is no longer complementary and the enzyme is denatured. The primary structure is peptide-bonded and survives; enzymes are catalysts and are not consumed.

What does the induced fit model propose?

• AThe active site moulds around the substrate as it binds✓
• BThe substrate moulds around the active site as it binds
• CThe active site and substrate are permanently complementary
• DThe enzyme and substrate form a permanent covalent bond

Answer: Induced fit replaced lock-and-key: the active site is flexible and changes shape slightly as the substrate enters, straining the substrate's bonds and lowering activation energy. The substrate is not the part that adjusts.

Which element is found in all proteins but not in carbohydrates?

• APhosphorus, present in the amine group
• BSulfur, present in every amino acid
• CNitrogen, present in the amine group✓
• DOxygen, present in the carboxyl group

Answer: Every amino acid carries an amine group, so nitrogen is always present. Carbohydrates and lipids contain only carbon, hydrogen and oxygen. Sulfur appears only in some R groups, and phosphorus belongs to nucleotides and phospholipids.

Why does a competitive inhibitor's effect lessen at high substrate concentration?

• ASubstrate binds the inhibitor and removes it
• BSubstrate raises the temperature of the reaction
• CSubstrate changes the shape of the active site
• DSubstrate outcompetes it for the active site✓

Answer: A competitive inhibitor is a similar shape to the substrate and occupies the active site. Raising substrate concentration means the active sites are far more likely to meet substrate than inhibitor, so the maximum rate is eventually reached. A non-competitive inhibitor cannot be outcompeted this way.

Why is water able to act as a solvent for ions?

• AIt is non-polar, so it surrounds and separates charged particles
• BIt has a high specific heat capacity, so it dissolves charged particles
• CIt has a high latent heat of vaporisation, so it dissolves charged particles
• DIt is polar, so it surrounds and separates charged particles✓

Answer: Uneven sharing of electrons leaves oxygen slightly negative and hydrogen slightly positive, so water molecules cluster around ions and pull them into solution. High specific heat capacity and latent heat are genuine properties of water but explain temperature buffering and cooling, not dissolving.