Topic 2: Biological molecules

 

Testing for biological molecules

Students should be able to:

1) Describe and carry out the Benedict’s test for reducing sugars, the iodine test for starch, the emulsion test for lipids and the biuret test for proteins

2) Describe and carry out a semi-quantitative Benedict’s test on a reducing sugar solution by standardising the test and using the results (time to first colour change or comparison to colour standards) to estimate the concentration

3) Describe and carry out a test to identify the presence of non-reducing sugars, using acid hydrolysis and Benedict’s solution

Benedict's test for reducing sugars:

Benedict's reagent is an alkaline solution containing copper(II) sulfate (CuSO₄), sodium carbonate (Na₂CO₃) to create basic conditions and sodium citrate as a complexing agent to keep Cu²⁺ in solution. Reducing sugars in their open chain forms have aldehyde or ketone groups that can reduce Cu²⁺ to Cu⁺. Two Cu⁺ ions form copper(I) oxide (Cu₂O), an insoluble brick‑red precipitate.
 

Detailed procedure (qualitative)

1. Label boiling tubes for standards and samples and set up a water bath to boil (100 °C).

2. Pipette 2.0 cm³ of the sample solution into a clean boiling tube.

3. Add 2.0 cm³ Benedict's reagent (use the same batch across all tubes).

4. Mix gently and place in the boiling water bath for exactly 3 minutes. Start a stopwatch when tubes are immersed.

5. Remove from bath, allow to cool slightly, and observe the colour and any precipitate.
    

Expected colours and interpretation

• Blue: no reducing sugar detected.

• Green/yellow: low concentration.

• Orange: moderate.

• Brick‑red precipitate: high concentration.
   

Semi‑quantitative method — time to first colour change

• Prepare glucose standards: 0.1, 0.2, 0.5, 1.0, 2.0 g dm⁻³ (prepare by weighing accurately and making up to volume). For each standard and the unknown, pipette 2.0 cm³ sample + 2.0 cm³ Benedict's into labelled tubes. Immerse all tubes simultaneously in a pre‑boiled water bath and start the timer. Record the time at which each tube first shows any colour change from blue (green or beyond). Take three repeats and calculate mean times for each concentration. Plot concentration (x) vs time (y) and interpolate for the unknown. Include negative (water) and positive (known glucose) controls.
  
   

 

Non‑reducing sugar test (acid hydrolysis + Benedict's):

A disaccharide like sucrose has a glycosidic bond between the anomeric carbons of its constituent monosaccharides, preventing the open‑chain aldehyde/ketone required for reducing activity. Hydrolysis with dilute acid breaks the glycosidic bond producing monosaccharides (glucose + fructose) that are reducing.
 

Procedure 

1. Confirm negative Benedict's on the original sample.

2. Pipette 2.0 cm³ of sample into a test tube and add 2.0 cm³ of 0.1 M HCl.

3. Heat gently in a water bath (not directly over flame) for 3–5 min.

4. Cool and carefully neutralise by adding small amounts of solid sodium hydrogen carbonate until fizzing (CO₂) stops and test pH to near neutral (pH 7–8) using universal indicator paper.

5. Perform Benedict's test on the neutralised mixture (2.0 cm³ hydrolysate + 2.0 cm³ Benedict's) and heat as usual.

Interpretation.

• If the original was negative and the hydrolysed sample becomes positive, a non‑reducing sugar (such as sucrose) was present.

• If still negative, either no sugar was present or sample concentration was too low.
   

 

Iodine test for starch:

Chemical basis Iodine (I₂) in potassium iodide forms a triiodide ion I₃⁻ that slips into the helical amylose structure forming a charge‑transfer complex that absorbs visible light strongly, producing a blue‑black colour. Amylopectin gives a reddish colour in some conditions due to different complex geometry but classically the test is described as blue‑black for starch.

Procedure (solid and solution samples).

1. For solids: place a small crumb or sample on a white tile and add 1–2 drops of iodine solution; observe.

2. For solutions: place 2.0 cm³ sample in a test tube and add 1–2 drops of iodine solution; mix gently and observe.
   
   Emulsion test for lipids (ethanol method):

Ethanol emulsion.

1. Add 2.0 cm³ absolute ethanol to a small piece of sample or 0.5 cm³ liquid sample, mix to dissolve lipids.

2. Add 2.0 cm³ water and shake. A milky emulsion indicates lipids.

Limitations.

• Emulsion test is qualitative; strongly coloured samples or turbidity from other suspended solids can confuse results. Use blanks and controls.
   

 

Biuret test for proteins:

In alkaline conditions copper(II) ions (Cu²⁺) form a coordination complex with lone pair electrons on peptide bond nitrogen atoms (–CONH–). This complex is violet due to d–d electronic transitions in the copper ion influenced by ligand field.
 

Procedure

1. Make sample alkaline by adding an equal volume of 1% (w/v) NaOH.

2. Add 1–2 drops of 0.05% copper(II) sulfate solution (or add 1 volume of Biuret reagent).

3. Gently mix and allow 2 minutes for full development of colour. Record as violet (positive) or blue (negative).
    

Interfering substances

• Free amino acids give weaker positive results; the test is most sensitive to polypeptides with multiple peptide bonds.
  
   

                                                                                              CARBOHYDRATES AND LIPIDS

Students should be able to:

1) Describe and draw the ring forms of α-glucose and β-glucose

2) Define the terms monomer, polymer, macromolecule, monosaccharide, disaccharide and polysaccharide

3) State the role of covalent bonds in joining smaller molecules together to form polymers

4) State that glucose, fructose and maltose are reducing sugars and that sucrose is a non-reducing sugar

5) Describe the formation of a glycosidic bond by condensation, with reference to disaccharides, including sucrose, and polysaccharides

6) Describe the breakage of a glycosidic bond in polysaccharides and disaccharides by hydrolysis, with reference to the non-reducing sugar test

7) Describe the molecular structure of the polysaccharides starch (amylose and amylopectin) and glycogen and relate their structures to their functions in living organisms

8) Describe the molecular structure of the polysaccharide cellulose and outline how the arrangement of cellulose molecules contributes to the function of plant cell walls

9) State that triglycerides are non-polar hydrophobic molecules and describe the molecular structure of triglycerides with reference to fatty acids (saturated and unsaturated), glycerol and the formation of ester bonds 10) Relate the molecular structure of triglycerides to their functions in living organisms

11) Describe the molecular structure of phospholipids with reference to their hydrophilic (polar) phosphate heads and hydrophobic (non-polar) fatty acid tails

​

2.2.1 Ring forms of α‑glucose and β‑glucose

Glucose predominantly exists in solution as a cyclic hemiacetal formed by reaction between the hydroxyl group on carbon‑5 and the aldehyde on carbon‑1, producing a six‑membered pyranose ring and creating a new stereocentre at the anomeric carbon so that the hydroxyl on C1 can be either below the plane of the ring (α‑D‑glucose) or above the plane (β‑D‑glucose); students should be able to draw these Haworth ring structures, label all six carbon atoms (C1–C6), and explain that the different configuration at the anomeric carbon affects the orientation of glycosidic bonds formed during condensation reactions — for example α‑1,4 glycosidic bonds give rise to coiled, helical chains in amylose while β‑1,4 bonds give straight chains in cellulose — and therefore that the seemingly small stereochemical difference has profound consequences for polymer structure and biological function. For top marks include the mechanistic note that ring formation is an equilibrium process and that the anomeric carbon is the reactive centre in mutarotation and glycosidic bond formation.
 

2.2.2 Definitions: monomer, polymer, macromolecule, monosaccharide, disaccharide, polysaccharide

A precise set of definitions is required: a monomer is a simple molecular unit capable of joining with identical or different units by covalent bonds to form a polymer (for example glucose is the monomer for starch and cellulose); a polymer is a large molecule composed of repeating monomeric units (for example polysaccharides like starch and cellulose); a macromolecule is any very large molecule often formed by polymerisation of smaller subunits (for example polysaccharides and proteins); a monosaccharide is the simplest carbohydrate that cannot be hydrolysed into smaller sugars (examples include glucose and fructose); a disaccharide consists of two monosaccharides joined by a glycosidic bond (examples include maltose and sucrose); and a polysaccharide is a polymer of many monosaccharide units such as starch, glycogen or cellulose — an exam answer that includes these definitions with a brief example for each term demonstrates both understanding and ability to apply definitions to biological molecules.
 

2.2.3 Role of covalent bonds in joining smaller molecules to form polymers

Covalent bonds are the strong chemical links that join smaller molecules into polymers and in carbohydrates these bonds are glycosidic (ether‑type) linkages formed by condensation reactions in which a hydroxyl group from one monosaccharide reacts with a hydroxyl group from another releasing a molecule of water and creating an O‑glycosidic bond; in biological systems such polymerisation is enzyme‑catalysed and often uses activated intermediates (for example UDP‑glucose) to make the overall process energetically favourable, while the reverse reaction — hydrolysis — adds water to break the glycosidic bond and is catalysed by specific hydrolase enzymes. For full marks students should explicitly link the covalent nature of the bond to polymer stability and how enzymes control formation and hydrolysis in vivo.
 

2.2.4 Which sugars are reducing: glucose, fructose, maltose vs sucrose

Glucose, fructose and maltose are reducing sugars because they can provide reducing equivalents under Benedict's test conditions: glucose is an aldose that forms an open‑chain aldehyde capable of reducing Cu²⁺, maltose retains a free anomeric carbon at one end and therefore also behaves as a reducing sugar, and fructose although a ketose can tautomerise under alkaline conditions to a form that reduces copper reagents — conversely sucrose is non‑reducing because its glycosidic bond joins the anomeric carbons of glucose and fructose (an α‑1→β‑2 linkage) so neither sugar unit has a free anomeric carbon available to form the open‑chain reactive aldehyde/ketone until the glycosidic bond is hydrolysed; a high‑scoring explanation links this structural description directly to the chemical basis of the Benedict's test and provides the example of sucrose becoming reducing after acid hydrolysis.
 

2.2.5 Formation of a glycosidic bond by condensation (disaccharides, including sucrose, and polysaccharides)

A glycosidic bond forms by a condensation (dehydration) reaction in which the hydroxyl group on the anomeric carbon of one monosaccharide reacts with the hydroxyl on another monosaccharide, releasing a molecule of water and producing an O‑glycosidic linkage whose designation (for example α‑1,4 or α‑1,6 or β‑1,4) indicates the carbons involved and the stereochemistry at the anomeric carbon; when two monosaccharides join this produces a disaccharide such as maltose (α‑1,4 glucose‑glucose) or sucrose (α‑1→β‑2 glucose‑fructose) and repeated condensation produces polysaccharides such as starch or cellulose, and a complete answer should note that in living cells these reactions are enzyme‑catalysed and often proceed via activated sugar intermediates. To score full credit students should name the bond type, describe the condensation mechanism briefly, and give specific examples linking bond type to the disaccharide or polysaccharide named.
 

2.2.6 Breakage of glycosidic bonds by hydrolysis with reference to non‑reducing sugar test

The breakage of glycosidic bonds is the reverse reaction — hydrolysis — in which water adds across the bond to reform the individual monosaccharides and this is catalysed in living organisms by hydrolase enzymes (for example amylase, maltase, sucrase) while in the laboratory strong acids under heat can be used to hydrolyse glycosidic bonds; this chemical principle underpins the non‑reducing sugar test where sucrose, which does not reduce Benedict's reagent, is first hydrolysed by acid into glucose and fructose, then neutralised and retested with Benedict's reagent to reveal reducing monosaccharides — a full answer describes the chemical change and links the laboratory hydrolysis step explicitly to the change from a negative to a positive Trent result.
 

2.2.7 Molecular structure of starch (amylose and amylopectin) and relation to function

Starch is a plant polysaccharide composed primarily of two molecular components with complementary properties: amylose is a largely unbranched polymer of α‑1,4 linked glucose residues that adopts a helical conformation allowing dense packing and limiting osmotic effect inside cells, while amylopectin is a highly branched polymer containing α‑1,4 linked chains and α‑1,6 linkages at branch points approximately every 24–30 residues which produces a bushy molecule with many terminal residues that can be rapidly acted on by enzymes to release glucose; these structural features explain function — amylose provides compact energy storage while amylopectin’s branched architecture ensures rapid mobilisation of glucose by providing many non‑reducing ends accessible to enzymes, and a full answer connects insolubility, compactness, and enzymatic accessibility to the biological role of starch as an efficient plant storage carbohydrate.
 

2.2.8 Molecular structure of glycogen and relation to function

Glycogen is the principal short‑term carbohydrate store in animals and fungi and is structurally similar to amylopectin but with a much higher frequency of α‑1,6 branch points (approximately every 8–12 residues), resulting in a highly branched, compact molecule that provides a very large number of non‑reducing ends where glycogen phosphorylase and debranching enzymes can act simultaneously to liberate glucose rapidly; this highly branched structure therefore permits fast release of glucose to meet sudden energy demands in tissues such as muscle and allows dense, osmotically inert storage in liver and muscle cells — an examination response should explicitly relate the branching frequency to the need for rapid mobilisation and briefly mention the enzymatic mechanisms of breakdown.
 

2.2.9 Molecular structure of cellulose and how arrangement contributes to plant cell wall function

Cellulose is a linear homopolymer of β‑D‑glucose units linked by β‑1,4 glycosidic bonds which result in each successive glucose being rotated 180° relative to its neighbour, producing straight chains that align in parallel and form an extensive network of hydrogen bonds both within and between chains to create microfibrils; these microfibrils aggregate into fibres that provide enormous tensile strength and rigidity important for plant cell walls, resisting internal turgor pressure and providing structural support for the plant body — the answer should emphasise how the β‑linkage geometry enables a straight, tightly hydrogen‑bonded structure and why this confers the mechanical and insoluble properties required for a structural polysaccharide. Top answers mention that most animals lack cellulases and therefore cannot digest cellulose, linking structure to ecological and physiological consequences.

  
2.2.10 Triglycerides: non‑polar hydrophobic nature and molecular structure with reference to fatty acids and glycerol and ester bonds

Triglycerides are non‑polar hydrophobic molecules composed of a glycerol backbone esterified to three fatty acid chains via ester bonds; glycerol contributes three hydroxyl groups that condense with the carboxyl groups of fatty acids to form ester linkages during synthesis, and the fatty acids themselves vary in chain length and degree of unsaturation — saturated fatty acids contain no carbon‑carbon double bonds and therefore pack tightly, while unsaturated fatty acids contain one or more C=C bonds (typically cis configuration in biological systems) which introduce kinks that prevent tight packing and lower melting points — because the long hydrocarbon chains are non‑polar triglycerides are insoluble in water, making them ideal for compact, long‑term energy storage, and the exam answer should describe both the chemical structure (glycerol + three fatty acids, ester bonds) and the physical chemical consequence (hydrophobicity) that explains their storage role.
 

2.2.10 — Relating triglyceride structure to function
A triglyceride is formed by esterification of three fatty acids to a glycerol backbone, producing a molecule dominated by long non-polar hydrocarbon chains; this non-polarity makes triglycerides hydrophobic and chemically inert in aqueous environments so they can be stored as dense, osmotically inactive droplets (e.g. adipose tissue, seed oils) without disturbing cell water potential. The long, highly reduced fatty-acid chains give triglycerides a very high energy density when oxidised, making them ideal long-term energy stores, and their chemical stability means they are metabolically inactive until lipases hydrolyse the ester bonds to release fatty acids for β-oxidation. Variation in chain saturation affects physical state and mobilisability (saturated fats pack tightly and are less fluid; unsaturated fats are more readily mobilised), and the low density and insulating properties of triglyceride deposits also provide buoyancy and thermal/mechanical protection; oxidation additionally yields metabolic water, useful in arid environments.

 

2.2.11 — Molecular structure of phospholipids
A phospholipid consists of a glycerol backbone with two fatty acid tails (non-polar, hydrophobic) and a phosphate-containing head group (polar, hydrophilic), making the molecule amphipathic. In water amphipathic phospholipids spontaneously arrange so heads face the aqueous phases and tails face inward, forming bilayers whose hydrophobic core acts as a selective barrier — this bilayer is the structural basis of biological membranes, providing compartmentalisation, a matrix for membrane proteins, and selective permeability. Tail length and degree of unsaturation control membrane fluidity (unsaturated tails ↑ fluidity), and sterols (e.g. cholesterol) further modulate packing and permeability; phosphate head groups also provide sites for protein binding and signalling.
 

                                                                           PROTEINS

Students should be able to:

1) describe and draw the general structure of an amino acid and the formation and breakage of a peptide bond

2) explain the meaning of the terms primary structure, secondary structure, tertiary structure and quaternary structure of proteins

3) describe the types of interaction that hold protein molecules in shape

4) state that globular proteins are generally soluble and have physiological roles and fibrous proteins are generally insoluble and have structural roles

5) describe the structure of a molecule of haemoglobin as an example of a globular protein, including the formation of its quaternary structure from two alpha (α) chains (α–globin), two beta (β) chains (β–globin) and a haem group

6) relate the structure of haemoglobin to its function, including the importance of iron in the haem group

7) describe the structure of a molecule of collagen as an example of a fibrous protein, and the arrangement of collagen molecules to form collagen fibres

8) relate the structures of collagen molecules and collagen fibres to their function

​

2.3.1 General structure of an amino acid and formation/breakage of a peptide bond

All standard amino acids share a common structural backbone consisting of an amino group (–NH₂), an alpha carbon bearing a hydrogen and a variable side chain (R group), and a carboxyl group (–COOH) so that the general formula can be written as H₂N–CH(R)–COOH; this arrangement makes the alpha carbon chiral in all amino acids except glycine and explains why amino acids have distinct chemical properties dependent on their R groups (such as polarity, charge and reactivity); peptide bonds form when the carboxyl group of one amino acid reacts with the amino group of another in a condensation (dehydration) reaction that releases a molecule of water and creates an amide linkage (–CO–NH–) called a peptide bond, which is planar and has partial double‑bond character due to resonance, restricting rotation and influencing secondary structure — the reverse process, hydrolysis, breaks peptide bonds by adding water across the bond and is catalysed in living systems by proteases. A top exam answer draws or describes the dipeptide showing the peptide bond explicitly and explains both formation and hydrolytic cleavage including the fact that peptide bonds are stabilised by resonance.
 

2.3.2 Primary, secondary, tertiary and quaternary structures of proteins

The primary structure of a protein is its linear sequence of amino acids linked by peptide bonds and this sequence encodes all information necessary for higher levels of structure because the chemical properties of side chains determine folding; secondary structure refers to local regular conformations including the α‑helix (stabilised by hydrogen bonds between the carbonyl oxygen of residue i and the amide hydrogen of residue i+4) and the β‑pleated sheet (stabilised by hydrogen bonds between strands which may be parallel or antiparallel), both arising from backbone hydrogen bonding; tertiary structure is the three‑dimensional folding of the entire polypeptide chain into a compact, functional conformation resulting from interactions among side chains — hydrophobic interactions that bury non‑polar residues in the core, hydrogen bonds between polar side chains, ionic bonds (salt bridges) between oppositely charged residues, van der Waals interactions and covalent disulfide bonds between cysteine residues — and quaternary structure exists where multiple polypeptide chains (subunits) associate non‑covalently or with covalent crosslinks to form a functional protein complex, as exemplified by haemoglobin which comprises four subunits; an effective exam answer defines each level and links them—for example mutations that alter primary sequence can disrupt tertiary folding and thus abolish function.
 

2.3.3 Types of interactions that hold protein molecules in shape: hydrophobic interactions, hydrogen bonding, ionic bonding, covalent (disulfide) bonds

Proteins are stabilised by a variety of chemical interactions that operate at different strengths and length scales: hydrophobic interactions arise because non‑polar side chains cluster away from the aqueous environment, increasing the entropy of surrounding water and driving folding; hydrogen bonds form between backbone and side chain polar groups and stabilise secondary and tertiary structures though individually weak they are collectively important; ionic bonds or salt bridges form between positively and negatively charged side chains (for example lysine/arginine with aspartate/glutamate) and contribute to stabilisation particularly at surfaces or in less polar environments; and covalent disulfide bonds between cysteine residues provide strong, often irreversible crosslinks that stabilise extracellular proteins exposed to variable conditions — a full answer should explain the chemical basis of each interaction, relative strength and where in proteins they are most commonly found and include examples such as the role of disulfide bonds in stabilising insulin or extracellular domains of secreted proteins.
 

2.3.4 Globular vs fibrous proteins — solubility and physiological roles

Globular proteins are generally compact, roughly spherical molecules in which hydrophilic side chains are exposed on the surface interacting with water, making these proteins usually soluble and appropriate for dynamic physiological roles such as enzymatic catalysis, transport (for example haemoglobin) and regulation; by contrast fibrous proteins consist of long, extended polypeptide chains with repetitive sequences and strong cross‑linking that produce insoluble, tough fibres adapted for structural roles — collagen and keratin are classic examples — and a succinct full‑mark paragraph will contrast solubility, typical structure, and give representative functions for each class while noting that there are intermediate cases and that post‑translational modifications often dictate whether a protein is secreted and forms extracellular fibres.
 

2.3.5 Structure of haemoglobin: organization into α and β chains and haem group

Haemoglobin is a tetrameric globular protein made of two alpha (α) and two beta (β) polypeptide chains; each polypeptide folds into a globin domain that contains a pocket for the haem prosthetic group, which is a planar porphyrin ring coordinating a central iron ion (Fe²⁺) that can bind one molecule of oxygen non‑covalently; the iron is coordinated to four nitrogens within the porphyrin and to a proximal histidine residue of the globin chain, leaving a sixth coordination site available for reversible O₂ binding — the quaternary assembly and specific subunit interfaces transmit conformational changes upon O₂ binding across the tetramer, enabling cooperative binding behaviour. A full answer should describe the quaternary composition α₂β₂, the presence of the haem prosthetic group with Fe²⁺, and explain how reversible oxygen binding at the haem site is facilitated by the protein environment.
 

2.3.6 Relate the structure of haemoglobin to its function, including importance of iron in the haem group

Haemoglobin's structure is directly tied to its role in oxygen transport: the haem group's central Fe²⁺ provides the chemical site for reversible O₂ binding, and the protein environment around each haem both protects the iron from oxidation to Fe³⁺ (which would form methemoglobin incapable of binding oxygen) and modulates oxygen affinity through conformational changes; importantly the tetrameric quaternary structure produces cooperative binding — the binding of oxygen to one haem raises the affinity of the remaining haems via conformational shifts at subunit interfaces — producing a sigmoidal oxygen dissociation curve that allows efficient oxygen uptake in the lungs and release in peripheral tissues, and a high‑scoring answer will tie the chemical capability of Fe²⁺ to bind oxygen with the protein’s quaternary dynamics that create cooperativity and physiological O₂ delivery. Mentioning clinical implications briefly (for example CO binding or methemoglobinaemia) can show deeper understanding.
 

2.3.7 Structure of collagen molecule and arrangement to form collagen fibres

Collagen is a fibrous protein whose basic structural unit is tropocollagen — a right‑handed triple helix formed from three left‑handed polypeptide chains with a repeating Gly–X–Y sequence where glycine at every third position allows the tight packing of helices at the core and where X and Y are frequently proline and hydroxyproline; post‑translational hydroxylation of proline and lysine residues (a vitamin C‑dependent reaction) and subsequent covalent crosslinking between tropocollagen molecules produce the staggered arrays and covalently linked fibrils that aggregate into collagen fibres, and these highly ordered, crosslinked structures provide the tensile strength and insolubility required for connective tissues. A complete answer will detail the Gly–X–Y motif, the triple helix geometry, the role of hydroxyproline and hydroxylysine in stabilisation and crosslink formation, and the hierarchical assembly from molecules to fibrils and fibres.
 

2.3.8 Relate collagen structure to function

Collagen's mechanical role in tissues is explained by its molecular architecture: the cross links and tight triple helix packing produce a rigid, rod‑like tropocollagen molecule that, when staggered and covalently crosslinked into fibrils, resists tensile forces and prevents tissue stretching or rupture; the chemical stability provided by crosslinks and post‑translational modifications ensures durability in extracellular matrices while the insolubility and fibrous arrangement are ideally suited to structural roles such as tendons, ligaments, skin and bone matrix — the best exam answers also link to physiological consequences of defective collagen, such as scurvy (defective proline hydroxylation) or certain genetic connective tissue disorders, illustrating the importance of biochemical modifications for macroscopic function.
 

                                                                                          WATER

Students should be able to:

1) Explain how hydrogen bonding occurs between water molecules and relate the properties of water to its roles in living organisms, limited to solvent action, high specific heat capacity and latent heat of vaporisation

​

2.4.1 Hydrogen bonding between water molecules

Water is a polar molecule resulting from oxygen's greater electronegativity relative to hydrogen; the oxygen atom bears a partial negative charge while the hydrogen atoms bear partial positive charges, and these partial charges enable each water molecule to form hydrogen bonds — directional, electrostatic interactions in which the δ⁺ hydrogen of one water molecule is attracted to the δ⁻ oxygen of another — producing a dynamic network of hydrogen bonds that continually forms and breaks on short timescales, and these hydrogen bonds are the molecular origin of many bulk properties of water, including cohesion, high surface tension, high boiling and melting points relative to comparable small molecules, and the ability to stabilise dissolved ions and polar molecules through hydration shells. A thorough exam answer describes polarity, geometry and hydrogen bonding and explains how the resulting network yields macroscopic properties.
 

2.4.2 Water as a solvent (solvent action) and biological roles

Because of its polarity and high dielectric constant, water is an excellent solvent for ionic and many polar substances: ions and polar molecules are stabilised by hydration shells in which water molecules orient their partial charges towards solute charges, reducing electrostatic attractions between ions and thereby allowing dissolution and transport; in living organisms this solvent action enables metabolic reagents and products, ions and small polar molecules to move within the cytoplasm and extracellular fluids, allows enzyme‑substrate encounters to occur efficiently in solution, supports the establishment of ionic gradients across membranes necessary for nerve impulses and muscle contraction, and underpins transport processes such as blood plasma carrying glucose and minerals — an effective answer links the microscopic explanation (hydration shells, polarity) to multiple biological outcomes and mentions the importance of water’s high dielectric constant in reducing electrostatic interactions.
 

2.4.3 High specific heat capacity of water and biological significance

Water has a high specific heat capacity because a significant fraction of the energy supplied when heating water is absorbed in disrupting hydrogen bonds rather than immediately increasing kinetic energy, meaning that water requires a large energy input to raise its temperature (about 4.18 J g⁻¹ K⁻¹ at typical conditions); biologically this thermal buffering stabilises internal temperatures of organisms and the temperature of aquatic habitats, reducing thermal fluctuations that could denature enzymes or disrupt metabolic homeostasis, and it provides an environmental inertia that allows organisms to maintain biochemical processes within functional temperature ranges, so when answering an exam question students should connect the molecular basis (hydrogen bond disruption) to thermal stability and specific organismal advantages such as protection of enzyme function and moderation of climate in aquatic environments.
 

2.4.4 High latent heat of vaporisation and evaporative cooling

The high latent heat of vaporisation of water arises because molecules in the liquid phase are hydrogen‑bonded, and considerable energy is required to break these hydrogen bonds for molecules to escape into the gas phase; as a result the evaporation of water removes large amounts of heat per gram of water lost which is exploited biologically in evaporative cooling mechanisms — sweating in mammals removes heat effectively because the evaporating water extracts thermal energy from the skin and blood, and transpiration in plants cools leaves and assists the upward movement of water and dissolved minerals through the transpiration stream — students should therefore explicitly link hydrogen bonding to latent heat values and to efficient physiological cooling and transport mechanisms.