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    Biological molecules — OCR A-Level Biology

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    Biological molecules explained

    Biological molecules covers 20 specification points, including (a) how hydrogen bonding occurs between water… and (b) the concept of monomers and polymers and….

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    Open a point below for its explanation and revision guidance.

    (a) how hydrogen bonding occurs between water molecules, and relate this, and other properties of water, to the roles of water for living organisms

    Water is a polar molecule: the oxygen atom carries a partial negative charge (δ⁻) and each hydrogen a partial positive charge (δ⁺). The δ⁺ hydrogen of one molecule is attracted to the δ⁻ oxygen of a neighbouring molecule, forming a hydrogen bond. Each water molecule can hydrogen bond with up to four others, giving a network that explains water's properties. High specific heat capacity and high latent heat of vaporisation buffer temperature changes and allow cooling by sweating. Cohesion and surface tension support columns of water in xylem and allow insects to move on water. Being a solvent for ions and polar substances lets water act as a transport medium in blood plasma and as a medium for metabolic reactions. Ice is less dense than liquid water, so it floats and insulates water below.

    (b) the concept of monomers and polymers and the importance of condensation and hydrolysis reactions in a range of biological molecules

    A monomer is a small, repeating subunit; a polymer is a large molecule made of many monomers joined together. Examples include monosaccharides forming polysaccharides, amino acids forming polypeptides and proteins, and nucleotides forming nucleic acids. Monomers join by condensation reactions, which form a chemical bond and release one molecule of water per bond formed. Polymers are broken down by hydrolysis reactions, which use a molecule of water to break the bond and release the monomers. These reactions are central to building storage and structural molecules (such as starch, glycogen and cellulose) and to digesting food so that soluble products can be absorbed. They also allow polymers to be assembled and dismantled as needed, for example during protein synthesis and digestion.

    (c) the chemical elements that make up biological molecules

    Biological molecules are built from a small set of chemical elements. The bulk of every carbohydrate, lipid, protein and nucleic acid is carbon, hydrogen and oxygen; proteins and nucleic acids additionally contain nitrogen, and nucleic acids also contain phosphorus. Sulphur appears in some amino acids, so it is present in certain proteins. A useful check is to link each element to a molecule class: C, H and O in all four classes; N in proteins and nucleic acids; P in nucleic acids (and in phospholipids); S in some proteins. In a multiple-choice question you may be asked which element is absent from a named class, or which class contains a named element. Read the option set carefully, eliminate options that add an element to a class that lacks it, and confirm the remaining choice against the class you were given.

    (d) the ring structure and properties of glucose as an example of a hexose monosaccharide and the structure of ribose as an example of a pentose monosaccharide

    Glucose and ribose are both sugars built from carbon, hydrogen and oxygen, but they differ in carbon number and in the ring they form. Glucose is a hexose: six carbons, molecular formula C₆H₁₂O₆. In solution its open chain closes so that carbon 1 (the aldehyde carbon) bonds to the oxygen on carbon 5, giving a six-membered pyranose ring in which five carbons and one oxygen form the ring and carbon 6 sits outside it. The ring closes in two forms, alpha and beta, differing in the position of the –OH on carbon 1. Ribose is a pentose: five carbons, formula C₅H₁₀O₅. Its ring uses four carbons plus one oxygen, a five-membered furanose ring. Both are reducing sugars with free hydroxyl groups, soluble because of hydrogen bonding to water, and glucose is the respiratory substrate.

    (e) the synthesis and breakdown of a disaccharide and polysaccharide by the formation and breakage of glycosidic bonds

    A condensation reaction joins two monosaccharides: the hydroxyl group of one and the hydrogen of the other combine, water is released, and a glycosidic bond forms. Two monosaccharides give a disaccharide, for example glucose plus fructose forming sucrose; many monosaccharides joined this way give a polysaccharide such as starch or glycogen. The reverse process is hydrolysis: water is added and the glycosidic bond breaks, releasing the monosaccharide units. In the body, enzymes such as amylase and maltase catalyse these hydrolyses; in a test tube, heating with dilute acid achieves the same breakage. Each bond formed or broken releases or consumes one water molecule, so a chain of n residues joined by n−1 glycosidic bonds needs n−1 water molecules to hydrolyse completely.

    (f) the structure of starch (amylose and amylopectin), glycogen and cellulose molecules

    All three are polymers of glucose but differ in bonding and chain shape. Starch contains amylose, an unbranched helix of α-glucose joined by 1,4-glycosidic bonds, and amylopectin, a branched chain with 1,4 links and 1,6 branches. Glycogen is also α-glucose with 1,4 links and more frequent 1,6 branches, giving a compact store. Cellulose is built from β-glucose; adjacent units are inverted, so each chain is straight and neighbouring chains are held by hydrogen bonds into strong microfibrils. These structural differences explain function: branching gives many ends for rapid hydrolysis, helices pack densely for storage, and straight hydrogen-bonded chains give strength to plant cell walls.

    (g) how the structures and properties of glucose, starch, glycogen and cellulose molecules relate to their functions in living organisms

    Glucose is a hexose monosaccharide; its soluble, reactive nature makes it the respiratory substrate transported in blood. Two glucose molecules condense to form maltose, and many form starch or glycogen. Starch is a compact, branched or helical polymer of α-glucose stored in plant cells; its insolubility and coiling make it a good storage carbohydrate. Glycogen is a more highly branched polymer of α-glucose stored in animal cells, giving rapid glucose release. Cellulose is a straight, unbranched polymer of β-glucose; alternating inversion lets chains form hydrogen-bonded microfibrils that give plant cell walls tensile strength. Relate each structural feature to the function it enables, and distinguish α- from β-glucose by the position of the hydroxyl group on carbon 1.

    (h) the structure of a triglyceride and a phospholipid as examples of macromolecules

    A triglyceride forms when one glycerol molecule condenses with three fatty acid molecules, releasing three water molecules and forming three ester bonds. The fatty acids may be saturated or unsaturated, and the triglyceride is a macromolecule used for energy storage because it is compact and insoluble. A phospholipid forms when one glycerol condenses with two fatty acids and one phosphate group, so it has a hydrophilic phosphate head and two hydrophobic fatty acid tails. This amphipathic structure lets phospholipids form bilayers in membranes. Recognise both as macromolecules built by condensation, and distinguish them by the third glycerol position: a fatty acid in a triglyceride, a phosphate group in a phospholipid.

    (i) the synthesis and breakdown of triglycerides by the formation and breakage of ester bonds between fatty acids and glycerol

    Triglycerides are built when one glycerol molecule bonds with three fatty acid molecules. Each bond forms by a condensation reaction: the hydroxyl group of glycerol reacts with the carboxyl group of a fatty acid, a water molecule is released, and an ester bond (—COO—) links the two. Repeating this three times gives a triglyceride plus three water molecules. Breakdown is the reverse: hydrolysis adds water across each ester bond, splitting the triglyceride back into glycerol and three fatty acids. This happens during digestion, catalysed by lipase, and in cells when fatty acids are respired. Learn the direction of each reaction, the number of water molecules involved and the location of the ester bond, then apply this to unfamiliar diagrams.

    (j) how the properties of triglyceride, phospholipid and cholesterol molecules relate to their functions in living organisms

    Each lipid's function follows from its structure. Triglycerides are non-polar and insoluble, store energy densely and provide insulation and buoyancy; their many carbon–hydrogen bonds release energy on oxidation. Phospholipids have a hydrophilic phosphate head and two hydrophobic fatty acid tails, so in water they form bilayers with tails inward, creating a selectively permeable barrier for membranes. Cholesterol is a small, rigid, amphipathic steroid: its hydrophobic region sits among fatty acid tails while its hydroxyl group faces the aqueous surface, restricting phospholipid movement to regulate membrane fluidity and stability. Link each structural feature to a named function and avoid describing properties without stating the function they enable.

    (k) the general structure of an amino acid

    An amino acid is built around a central (alpha) carbon atom. Attached to this carbon are four groups: an amine group (—NH₂), a carboxyl group (—COOH), a hydrogen atom (—H) and a variable R group (side chain). The amine and carboxyl groups are common to every amino acid; it is the R group that differs between the twenty or so standard amino acids and gives each its distinct properties, such as charge or polarity. Because the same carbon carries both an acidic carboxyl group and a basic amine group, amino acids can act as zwitterions and buffer solutions. In an MCQ you may be asked to identify a component, spot a wrongly drawn structure, or recognise that the R group is the only part that varies. Learn the four groups and their positions on the central carbon.

    (l) the synthesis and breakdown of dipeptides and polypeptides, by the formation and breakage of peptide bonds

    A dipeptide forms when the carboxyl group of one amino acid reacts with the amine group of another in a condensation reaction, releasing one molecule of water and creating a peptide bond between the two residues. Repeating this condensation links many amino acids into a polypeptide, with a peptide bond at each join. Breakdown is the reverse: hydrolysis adds water across each peptide bond, splitting the chain back into amino acids. In an MCQ you may be asked what is released during condensation, what is added during hydrolysis, or where the peptide bond forms. Remember that the bond joins the carbon of one carboxyl group to the nitrogen of the next amine group, and that the number of water molecules released equals the number of peptide bonds formed.

    (m) the levels of protein structure

    Proteins are built from amino acids joined by peptide bonds, and their shape arises in a hierarchy of levels. The primary structure is the sequence and number of amino acids in the polypeptide chain, held by covalent peptide bonds. The secondary structure is the local folding of the backbone into α-helices and β-pleated sheets, stabilised by hydrogen bonds between the C=O of one amino acid and the N–H of another. The tertiary structure is the further three-dimensional folding of the whole chain, held by hydrogen bonds, ionic bonds, hydrophobic interactions and disulfide bridges between R groups. The quaternary structure exists when two or more polypeptide chains associate, as in haemoglobin. A mutation that changes one amino acid alters the primary structure and can change the tertiary shape and function.

    (n) the structure and function of globular proteins including a conjugated protein

    Globular proteins fold so that hydrophobic residues cluster in the core and hydrophilic residues face the aqueous surroundings, giving a compact, roughly spherical shape that is soluble. The precise three-dimensional shape comes from the primary sequence and is held by hydrogen bonds, ionic bonds, disulfide bridges and hydrophobic interactions; this shape creates a specific active site or binding site, so function depends on structure. Examples include enzymes, haemoglobin and antibodies. A conjugated protein contains a non-protein prosthetic group. Haemoglobin is the standard example: four polypeptide chains each associated with a haem group containing an iron ion (Fe²⁺) that binds oxygen. The prosthetic group is essential to function, so removing it destroys oxygen transport.

    (o) the properties and functions of fibrous proteins

    Fibrous proteins are long, strand-like molecules whose amino acid chains run parallel along one axis and are held together by cross-links, so they are insoluble in water and structurally strong. Their repeating sequence and extensive hydrogen bonding and, where present, disulfide bridges give tensile strength and resistance to stretching. Collagen is the classic example: three polypeptide chains wound into a triple helix, cross-linked in fibrils, giving high tensile strength in tendons, skin and blood-vessel walls. Keratin has many cysteine residues forming disulfide bridges, giving hardness and waterproofing in hair, nails and skin. Elastin is extensible and recoils. Learn to link each structural feature to a named function and to contrast fibrous proteins with globular ones, which are compact and soluble.

    (p) the key inorganic ions that are involved in biological processes

    Teach the key inorganic ions required for OCR A Level Biology H420: hydrogen (H⁺), sodium (Na⁺), potassium (K⁺), calcium (Ca²⁺), magnesium (Mg²⁺), ammonium (NH₄⁺), nitrate (NO₃⁻), hydrogencarbonate (HCO₃⁻), chloride (Cl⁻), phosphate (PO₄³⁻) and hydroxide (OH⁻). For each ion, link its charge and size to a named biological role, e.g. H⁺ controls pH and enzyme activity; Na⁺ co-transports glucose in the ileum; K⁺ maintains resting potential in neurones; Ca²⁺ triggers muscle contraction and exocytosis; Mg²⁺ is central in chlorophyll; NH₄⁺ is a nitrogen source for plants; NO₃⁻ is absorbed for amino acid synthesis; HCO₃⁻ buffers blood and neutralises stomach acid in the duodenum; Cl⁻ balances charge and forms HCl in the stomach; PO₄³⁻ is in ATP, DNA and phospholipids; OH⁻ contributes to alkaline conditions.

    (q) how to carry out and interpret the results of the following chemical tests: • biuret test for proteins • Benedict’s test for reducing and non-reducing sugars • iodine test for starch • emulsion test for lipids

    Each test detects a specific molecule through a colour change. Biuret: add an equal volume of biuret solution (sodium hydroxide plus dilute copper(II) sulfate) to the sample; a purple colour indicates peptide bonds, so protein is present, while a blue colour is negative. Benedict's: add Benedict's reagent (blue copper(II) sulfate) and heat in a water bath; a brick-red precipitate shows a reducing sugar. For a non-reducing sugar, first boil with dilute hydrochloric acid to hydrolyse it, neutralise with sodium hydrogencarbonate, then repeat Benedict's; a positive result after hydrolysis but not before indicates a non-reducing sugar such as sucrose. Iodine: add iodine in potassium iodide solution; blue-black indicates starch, orange-brown is negative. Emulsion: add ethanol and shake, then pour into water; a cloudy white emulsion indicates lipid.

    (r) quantitative methods to determine the concentration of a chemical substance in a solution

    Quantitative methods measure how much of a substance is present, not merely whether it is present. A common approach is colorimetry: react the sample with a reagent that produces a coloured product, then measure absorbance with a colorimeter. Prepare a series of known concentrations (standard solutions), measure each absorbance, and plot a calibration curve of absorbance against concentration. Measure the absorbance of the unknown and read its concentration from the curve. Other quantitative methods include titration (for example determining reducing sugar concentration) and using a biosensor or a quantitative Benedict's series comparing precipitate with known standards. Reliable results require controlled volumes, consistent conditions and repeats to reduce error.

    (s) (i) the principles and uses of paper and thin layer chromatography to separate biological molecules / compounds

    Chromatography separates molecules by differential partitioning between a mobile phase and a stationary phase. In paper chromatography the stationary phase is water held in cellulose fibres; in thin layer chromatography (TLC) it is a thin layer of silica gel or cellulose on a glass or plastic plate. The solvent (mobile phase) moves up by capillary action, carrying dissolved molecules. Molecules that interact more strongly with the stationary phase travel more slowly, so each compound moves a characteristic distance. The result is expressed as an Rf value: Rf = distance travelled by the spot ÷ distance travelled by the solvent front. Rf is a ratio with no units and is always less than 1. Because Rf depends on the solvent and conditions, comparisons must use the same solvent and temperature.

    (ii) practical investigations to analyse biological solutions using paper or thin layer chromatography

    A practical chromatography investigation starts with a prepared sample: a small, concentrated spot of the biological solution is applied to the origin line, drawn in pencil, above the level of the solvent. The solvent is allowed to travel up the paper or TLC plate by capillary action. When the solvent front nears the top, the chromatogram is removed and the front is marked immediately in pencil. Spots may be colourless, so they are located using a locating agent such as ninhydrin for amino acids, or viewed under ultraviolet light. Each spot's distance from the origin is measured to its centre, and Rf is calculated as spot distance ÷ solvent-front distance. The sample is identified by comparing its Rf values with those of known standards run on the same chromatogram under identical conditions.

    Your focus

    1. Describe how hydrogen bonding occurs between water molecules, including the partial charges involved.
    2. Explain how named properties of water arise from hydrogen bonding and polarity.
    3. Relate properties of water to specific roles for living organisms, such as transport, cooling and insulation.
    Show all 61 objectives
    1. Define monomer and polymer and give biological examples of each.
    2. Explain the role of condensation reactions in joining monomers and releasing water.
    3. Explain the role of hydrolysis reactions in breaking polymers and using water, and relate both to biological processes.
    4. State the chemical elements present in carbohydrates, lipids, proteins and nucleic acids.
    5. Identify which element is present in one class of biological molecule but absent from another.
    6. Select the correct option describing the elemental composition of a named biological molecule.
    7. Identify glucose as a hexose and ribose as a pentose, giving their molecular formulae.
    8. Describe the ring structure of each sugar, stating which atoms form the ring.
    9. Explain how structural features give glucose and ribose their solubility and reducing properties.
    10. Describe condensation as the formation of a glycosidic bond with release of water.
    11. Describe hydrolysis as the breakage of a glycosidic bond with addition of water.
    12. Apply the n−1 bond relationship to a named polysaccharide chain.
    13. Compare the bonding and branching of amylose, amylopectin and glycogen.
    14. Explain how β-glucose produces straight cellulose chains held by hydrogen bonds.
    15. Relate structural features of each polymer to its biological role.
    16. Relate the solubility and reactivity of glucose to its role as a respiratory substrate.
    17. Explain how the structure of starch and glycogen suits storage in plants and animals respectively.
    18. Describe how the structure of cellulose provides strength in plant cell walls.
    19. Describe the subunits and bonds that make up a triglyceride.
    20. Describe the subunits and bonds that make up a phospholipid.
    21. Explain how the arrangement of hydrophilic and hydrophobic regions in a phospholipid suits membrane formation.
    22. Describe the condensation reaction that forms a triglyceride from glycerol and three fatty acids.
    23. Explain how hydrolysis breaks ester bonds to release glycerol and fatty acids.
    24. Identify ester bonds and the reacting groups in a triglyceride structure.
    25. Relate the properties of triglycerides to energy storage, insulation and protection.
    26. Explain how phospholipid structure produces a bilayer that acts as a membrane barrier.
    27. Describe how cholesterol's structure allows it to regulate membrane fluidity.
    28. Label the four groups attached to the central carbon of a general amino acid
    29. Distinguish the amine group from the carboxyl group by formula and position
    30. Explain why the R group makes amino acids differ from one another
    31. Describe condensation and hydrolysis in terms of water released or added
    32. Identify the peptide bond and the groups it joins
    33. Relate the number of amino acids to the number of peptide bonds in a chain
    34. Describe the four levels of protein structure and the bonds or interactions that stabilise each.
    35. Distinguish secondary from tertiary structure using correct bonding.
    36. Explain how a change in primary structure can alter tertiary structure and protein function.
    37. Describe the folded structure of globular proteins and the bonds that maintain it.
    38. Explain how the shape of a globular protein determines its function.
    39. Describe a conjugated protein, naming its prosthetic group and explaining its role in function.
    40. Describe the elongated, cross-linked structure of fibrous proteins.
    41. Explain how structural features of collagen, keratin and elastin produce their functions.
    42. Distinguish fibrous proteins from globular proteins in shape, solubility and role.
    43. State the chemical symbol and charge of each key inorganic ion.
    44. Describe one named biological role for each key inorganic ion.
    45. Explain how the charge or concentration of an ion relates to its function in a given process.
    46. Select the correct ion when given a description of a biological process.
    47. Describe how to carry out the biuret, Benedict's, iodine and emulsion tests.
    48. State the positive and negative results of each chemical test.
    49. Interpret observed results to identify the molecule present, including distinguishing reducing from non-reducing sugars.
    50. Explain what a quantitative method determines compared with a qualitative test.
    51. Describe how a calibration curve is produced and used to find an unknown concentration.
    52. Identify controls and repeats needed for reliable quantitative results.
    53. Describe the stationary and mobile phases in paper and thin layer chromatography.
    54. Explain how differential partitioning between phases separates biological molecules.
    55. Calculate and interpret Rf values and explain why conditions must be standardised for identification.
    56. Carry out a paper or thin layer chromatography investigation to separate components of a biological solution.
    57. Use a locating agent or ultraviolet light to detect separated spots and measure distances accurately.
    58. Calculate Rf values and identify components by comparison with known standards run under the same conditions.

    Biological molecules exam tips

    Marking Points
    • Hydrogen bonding arises because water is polar: oxygen is δ⁻ and hydrogen is δ⁺, so the δ⁺ hydrogen of one molecule is attracted to the δ⁻ oxygen of another.
    • Each water molecule can form hydrogen bonds with up to four neighbouring molecules, creating an extensive network.
    • High specific heat capacity means large energy changes cause only small temperature changes, so water buffers temperature in organisms and habitats.
    • High latent heat of vaporisation means much energy is needed to evaporate water, so sweating or transpiration cools organisms effectively.
    • Cohesion between water molecules and surface tension allow continuous columns in xylem and support organisms on the water surface.
    • Water is a good solvent for ions and polar molecules, so it acts as a transport medium (for example in blood plasma) and as a medium for metabolic reactions.
    • Ice is less dense than liquid water, so it floats and insulates the water beneath, helping organisms survive in cold conditions.
    • A monomer is a small repeating subunit; a polymer is a large molecule formed from many monomers joined together.
    • Condensation reactions join monomers, forming a bond and releasing one molecule of water per bond formed.
    • Hydrolysis reactions use water to break bonds in polymers, releasing the constituent monomers.
    • Examples include monosaccharides forming polysaccharides, amino acids forming polypeptides, and nucleotides forming nucleic acids.
    • Condensation builds storage and structural polymers such as starch, glycogen, cellulose and proteins.
    • Hydrolysis allows digestion of polymers into soluble monomers that can be absorbed, and allows polymers to be broken down when required.
    • Carbon, hydrogen and oxygen are present in carbohydrates, lipids, proteins and nucleic acids.
    • Nitrogen is present in proteins and nucleic acids but is not a component of carbohydrates or triglycerides.
    • Phosphorus is present in nucleic acids and in phospholipids.
    • Sulphur is present in some amino acids and therefore in some proteins.
    • Correctly matching an element to the class of biological molecule it occurs in, and rejecting an option that assigns an element to a class that lacks it.
    • Glucose is a hexose with six carbon atoms and formula C₆H₁₂O₆; ribose is a pentose with five carbons and formula C₅H₁₀O₅.
    • Glucose forms a six-membered ring containing five carbons and one oxygen (pyranose), with carbon 6 outside the ring.
    • Ribose forms a five-membered ring containing four carbons and one oxygen (furanose).
    • Both are soluble in water because their many hydroxyl groups hydrogen-bond with water molecules.
    • Glucose exists as alpha and beta ring forms differing in the orientation of the hydroxyl group on carbon 1.
    • Both are reducing sugars because the ring can open to expose a free aldehyde/ketone group that reduces other substances.
    • Condensation joins monosaccharides and releases one water molecule per glycosidic bond formed.
    • A disaccharide forms from two monosaccharides; a polysaccharide forms from many monosaccharides joined by glycosidic bonds.
    • Hydrolysis adds water and breaks glycosidic bonds, releasing monosaccharide units.
    • Enzymes such as amylase and maltase catalyse hydrolysis of glycosidic bonds in digestion.
    • The number of glycosidic bonds in a linear chain of n residues is n−1, so complete hydrolysis consumes n−1 water molecules.
    • Starch is made of amylose (unbranched helical α-glucose chains with 1,4 bonds) and amylopectin (branched with 1,6 bonds).
    • Glycogen is a branched polymer of α-glucose with 1,4 links and frequent 1,6 branches.
    • Cellulose is a polymer of β-glucose with alternating inverted units forming straight chains.
    • Hydrogen bonds between cellulose chains form strong microfibrils.
    • Branching increases the number of ends available for rapid hydrolysis, suiting energy storage.
    • Glucose is a hexose monosaccharide that is soluble and readily respired, so it can be transported and used as a respiratory substrate.
    • Starch is a polymer of α-glucose that is insoluble and often branched or coiled, making it a compact, osmotically inactive storage molecule in plants.
    • Glycogen is a highly branched polymer of α-glucose stored in animals, and its many branch ends allow rapid hydrolysis to release glucose.
    • Cellulose is an unbranched polymer of β-glucose; alternating inversion allows hydrogen bonding between chains to form strong microfibrils in cell walls.
    • Condensation joins monosaccharides with the loss of water, and hydrolysis reverses this; α-glucose and β-glucose differ in the position of the hydroxyl group on carbon 1.
    • A triglyceride consists of one glycerol molecule bonded to three fatty acid molecules by ester bonds formed in condensation reactions.
    • Three water molecules are released when one triglyceride is formed by condensation.
    • A phospholipid consists of one glycerol bonded to two fatty acids and one phosphate group.
    • The phosphate group gives a hydrophilic head, while the two fatty acid tails are hydrophobic, allowing phospholipids to form bilayers.
    • Both triglycerides and phospholipids are macromolecules assembled from smaller subunits by condensation.
    • Synthesis is a condensation reaction joining glycerol to three fatty acids, releasing one water molecule per bond formed.
    • Each ester bond forms between the hydroxyl group of glycerol and the carboxyl group of a fatty acid.
    • A triglyceride contains three ester bonds, so complete synthesis releases three water molecules.
    • Hydrolysis breaks ester bonds by adding water, regenerating glycerol and three fatty acids.
    • Lipase catalyses hydrolysis of triglycerides during digestion; the reverse condensation is also enzyme-controlled in cells.
    • Triglycerides are insoluble and non-polar, making them good energy stores and providing insulation and protection.
    • Triglycerides contain many carbon–hydrogen bonds, so oxidation releases large amounts of energy per gram.
    • Phospholipids are amphipathic, with a hydrophilic head and hydrophobic tails, so they form bilayers in water.
    • The phospholipid bilayer forms the selectively permeable barrier of cell membranes.
    • Cholesterol is amphipathic and rigid; it packs between phospholipid tails, regulating membrane fluidity and stability.
    • Identifies the central (alpha) carbon as the atom to which the four groups are bonded
    • Names the amine group (—NH₂) and the carboxyl group (—COOH) as the groups common to all amino acids
    • States that a hydrogen atom and a variable R group are also attached to the central carbon
    • Recognises that the R group (side chain) is the part that differs between different amino acids
    • Applies the structure to explain a property, such as the R group determining charge, polarity or reactivity
    • States that condensation joins the carboxyl group of one amino acid to the amine group of another, releasing water
    • Identifies the peptide bond as the link formed between the two amino acid residues
    • Explains that a dipeptide contains two amino acids joined by one peptide bond, while a polypeptide contains many
    • States that hydrolysis adds water to break peptide bonds, reversing condensation
    • Links the number of water molecules released or used to the number of peptide bonds formed or broken
    • Primary structure is the sequence (and number) of amino acids in the polypeptide chain, joined by peptide bonds.
    • Secondary structure is the local coiling or folding of the backbone into α-helices and β-pleated sheets, held by hydrogen bonds between backbone C=O and N–H groups.
    • Tertiary structure is the further three-dimensional folding of a single polypeptide, stabilised by hydrogen bonds, ionic bonds, hydrophobic interactions and disulfide bridges between R groups.
    • Quaternary structure is the association of two or more polypeptide chains (possibly with prosthetic groups) into a functional protein, as in haemoglobin with four chains.
    • A change in primary structure, such as a single substituted amino acid, can alter the folding and therefore the tertiary structure and function of the protein.
    • Globular proteins have a compact, roughly spherical shape with hydrophobic residues in the core and hydrophilic residues on the surface, making them soluble.
    • The tertiary structure is held by hydrogen bonds, ionic bonds, disulfide bridges and hydrophobic interactions, determined by the primary sequence.
    • The specific folded shape creates a precise active site or binding site, so structure relates directly to function.
    • A conjugated protein contains a non-protein prosthetic group permanently associated with the polypeptide.
    • In haemoglobin each of the four polypeptide chains is associated with a haem group containing an Fe²⁺ ion that binds oxygen.
    • The prosthetic group is required for function; without the haem group the protein cannot transport oxygen.
    • Fibrous proteins have long, parallel polypeptide chains forming extended strand-like structures rather than compact folded shapes.
    • Cross-links such as hydrogen bonds and disulfide bridges between chains give strength and insolubility in water.
    • Collagen consists of three polypeptide chains wound into a triple helix, providing tensile strength in tendons, skin and blood-vessel walls.
    • Keratin contains many cysteine residues whose disulfide bridges give hardness and waterproofing in hair, nails and skin.
    • Elastin is extensible and recoils, so it functions in tissues that must stretch and return, such as lung and artery walls.
    • Fibrous proteins are structural, whereas globular proteins are compact, soluble and often metabolic (for example enzymes).
    • Hydrogen ions (H⁺) determine pH; a higher H⁺ concentration lowers pH and can denature enzymes by disrupting hydrogen and ionic bonds.
    • Sodium ions (Na⁺) are involved in co-transport of glucose and amino acids across cell membranes, e.g. in the ileum, and in generating action potentials in neurones.
    • Potassium ions (K⁺) maintain the resting potential of neurones and are actively pumped into cells by the sodium–potassium pump.
    • Calcium ions (Ca²⁺) trigger muscle contraction by binding to troponin and stimulate exocytosis of neurotransmitters at synapses.
    • Magnesium ions (Mg²⁺) are the central ion in chlorophyll, essential for light absorption in photosynthesis.
    • Ammonium ions (NH₄⁺) are a source of nitrogen for plants, taken up by roots and used to synthesise amino acids and proteins.
    • Nitrate ions (NO₃⁻) are absorbed by plants and reduced to ammonium before incorporation into amino acids and nucleotides.
    • Hydrogencarbonate ions (HCO₃⁻) act as a buffer in blood and neutralise stomach acid in the duodenum, protecting the intestinal lining.
    • Chloride ions (Cl⁻) balance positive charges across membranes and are secreted into the stomach to form hydrochloric acid.
    • Phosphate ions (PO₄³⁻) are components of ATP, DNA, RNA and phospholipids, and are involved in phosphorylation reactions.
    • Hydroxide ions (OH⁻) contribute to alkaline conditions, e.g. in the small intestine, and are part of the water dissociation equilibrium.
    • Biuret test: add biuret solution to the sample; purple indicates protein (peptide bonds present), blue is a negative result.
    • Benedict's test for reducing sugar: add Benedict's reagent and heat in a water bath; a brick-red precipitate indicates a reducing sugar.
    • Non-reducing sugar: boil with dilute acid to hydrolyse, neutralise with sodium hydrogencarbonate, then repeat Benedict's; positive only after hydrolysis indicates a non-reducing sugar.
    • Iodine test: add iodine in potassium iodide solution; blue-black indicates starch, orange-brown is negative.
    • Emulsion test: add ethanol and shake, then pour into water; a cloudy white emulsion indicates lipid.
    • Interpretation requires comparing the observed colour or precipitate with the known positive and negative outcomes for each test.
    • Quantitative methods determine the amount or concentration of a substance, not just its presence.
    • Colorimetry: react the sample to produce a coloured product and measure absorbance with a colorimeter.
    • Prepare standard solutions of known concentration, measure their absorbance and plot a calibration curve.
    • Measure the unknown's absorbance and read its concentration from the calibration curve.
    • Other quantitative approaches include titration and comparison against a series of known standards.
    • Control variables such as volumes and conditions, and use repeats, to improve reliability.
    • Paper chromatography uses water held in cellulose as the stationary phase; TLC uses a thin layer of silica gel or cellulose on a plate as the stationary phase.
    • In both methods the mobile phase is a solvent that moves through the stationary phase, carrying the dissolved molecules with it.
    • Separation depends on differential partitioning: molecules with greater affinity for the stationary phase move more slowly than those with greater affinity for the mobile phase.
    • Rf = distance travelled by the spot ÷ distance travelled by the solvent front; it is a unitless ratio less than 1.
    • Rf values are only comparable when the same solvent, temperature and conditions are used, so known standards are run alongside samples.
    • Uses include separating and identifying amino acids, sugars and photosynthetic pigments in biological mixtures.
    • Apply a small, concentrated spot of the sample on a pencil origin line, keeping the spot above the solvent level so the sample does not dissolve into the reservoir.
    • Allow the solvent to move up the stationary phase by capillary action and mark the solvent front in pencil as soon as the run finishes.
    • Locate colourless spots using a locating agent such as ninhydrin for amino acids, or by viewing under ultraviolet light.
    • Measure from the origin line to the centre of each spot and to the solvent front, then calculate Rf = spot distance ÷ solvent-front distance.
    • Identify components by comparing sample Rf values with those of known standards run on the same chromatogram under the same conditions.
    • Use the results to analyse the biological solution, for example identifying which amino acids or sugars are present in a mixture.
    Examiner Tips
    • 💡Read the question stem to see whether it asks about the bonding itself or about a property linked to a role, and answer that part specifically.
    • 💡When linking a property to a role, name both the property and the living process it supports, for example cohesion supporting the transpiration stream.
    • 💡Eliminate options that describe covalent bonding or non-polar water before choosing between the remaining plausible properties.
    • 💡Check whether the question asks about building (condensation) or breaking (hydrolysis) before selecting an option.
    • 💡Link each reaction to a named biological example, such as amino acids forming a polypeptide by condensation.
    • 💡Watch for options that confuse the direction of water movement in the two reactions.
    • 💡Build a quick grid of molecule class against C, H, O, N, P, S and rehearse it before the exam.
    • 💡When an option adds an element to a class, test it against a specific example such as glucose or a triglyceride.
    • 💡Eliminate two clearly wrong options first, then decide between the remaining pair using the element you are most confident about.
    • 💡Learn the two molecular formulae and carbon counts so you can distinguish hexose from pentose quickly.
    • 💡Sketch each ring and label which atoms form the ring and which lie outside it.
    • 💡Link solubility and reducing behaviour to specific structural features (hydroxyl groups, ring opening) rather than stating them without reason.
    • 💡State clearly that water is released in condensation and used in hydrolysis, naming the bond each time.
    • 💡Link the reaction to a named example, such as glucose and fructose forming sucrose, to show understanding.
    • 💡When counting bonds, remember a chain of n units has n−1 glycosidic bonds.
    • 💡Name the glycosidic bond types (1,4 and 1,6) when comparing the polymers.
    • 💡Link each structural feature to a function, such as branching for rapid energy release.
    • 💡Contrast α-glucose and β-glucose to explain why cellulose chains are straight.
    • 💡For each molecule, state the structural feature and then the function it enables, rather than listing features alone.
    • 💡Use precise terms such as α-glucose, β-glucose, condensation, hydrolysis and microfibril.
    • 💡When comparing storage polysaccharides, focus on branching and its effect on the rate of glucose release.
    • 💡Count the subunits and bonds explicitly: glycerol plus three fatty acids gives three ester bonds in a triglyceride.
    • 💡Use the terms hydrophilic and hydrophobic accurately when describing the phospholipid head and tails.
    • 💡Link the amphipathic nature of phospholipids to bilayer formation in membranes.
    • 💡Count bonds and water molecules explicitly: three fatty acids means three ester bonds and three water molecules in condensation.
    • 💡Use the words condensation and hydrolysis precisely, matching each to synthesis or breakdown.
    • 💡When shown a structure, identify the ester linkage and state which groups reacted to form it.
    • 💡Structure the answer as feature, then property, then function for each lipid type.
    • 💡Use precise terms such as amphipathic, hydrophilic, hydrophobic and selectively permeable.
    • 💡Name the organism context, for example energy storage in adipose tissue or membrane fluidity in animal cells.
    • 💡Memorise the four groups attached to the central carbon and their positions before attempting structure questions.
    • 💡When checking a drawn structure, verify that the amine, carboxyl, hydrogen and R groups all attach to one central carbon.
    • 💡Read each option fully; distractors often swap the amine and carboxyl groups or move a group to the wrong atom.
    • 💡Link the term to the water: condensation removes water, hydrolysis adds water.
    • 💡Count peptide bonds as one fewer than the number of amino acids in a single chain.
    • 💡Check which groups are joined when identifying a peptide bond in a diagram.
    • 💡Learn each level with its bonds: peptide bonds for primary, backbone hydrogen bonds for secondary, R-group interactions for tertiary, and chain association for quaternary.
    • 💡Use haemoglobin as a concrete quaternary example and a single-chain enzyme such as lysozyme as a protein with no quaternary structure.
    • 💡When a question gives a mutation, trace the effect in order: changed amino acid sequence, altered R-group interactions, changed tertiary shape, changed function.
    • 💡Always link a named structural feature to a named function rather than listing bonds alone.
    • 💡For conjugated proteins, name the prosthetic group and state what it does, e.g. haem binds oxygen via Fe²⁺.
    • 💡Use haemoglobin as your worked example and be ready to explain why its quaternary structure and prosthetic groups matter.
    • 💡When a question names a fibrous protein, state one structural feature and then the function it produces, rather than listing functions alone.
    • 💡Use precise terms such as triple helix, disulfide bridge and tensile strength instead of vague words like strong or tough.
    • 💡If asked to compare, set fibrous against globular proteins on shape, solubility and role in one sentence each.
    • 💡Learn the ion, its charge and one precise biological role as a linked set; exam questions often ask for the ion when given a function.
    • 💡Use the context of the question to identify the ion: muscle contraction points to Ca²⁺, chlorophyll to Mg²⁺, resting potential to K⁺.
    • 💡Be precise with charges: write H⁺, Na⁺, K⁺, Ca²⁺, Mg²⁺, NH₄⁺, NO₃⁻, HCO₃⁻, Cl⁻, PO₄³⁻, OH⁻.
    • 💡For buffer questions, name hydrogencarbonate (HCO₃⁻) rather than a vague 'buffer ion'.
    • 💡If asked about plant mineral nutrition, distinguish between nitrate (NO₃⁻) and ammonium (NH₄⁺) as nitrogen sources.
    • 💡Learn the reagent, the condition (for example heating for Benedict's) and the positive colour for each test as a matched set.
    • 💡For non-reducing sugars, state the full sequence: hydrolyse with acid, neutralise, then repeat Benedict's.
    • 💡When interpreting, always state what the observed colour shows about the molecule present, not just the colour name.
    • 💡Describe the full colorimetry sequence: standards, calibration curve, then read the unknown.
    • 💡State clearly that a calibration curve links absorbance to known concentrations.
    • 💡Mention controlling variables and repeats when explaining how to obtain reliable quantitative results.
    • 💡Always state both phases explicitly: name the stationary phase and the mobile phase for the method described.
    • 💡When calculating Rf, show the two measured distances and divide spot distance by solvent-front distance, keeping the answer below 1.
    • 💡Link separation to relative affinity for the two phases rather than saying molecules are simply 'pulled up' at different speeds.
    • 💡For identification questions, mention comparing the sample Rf with Rf values of known standards run under the same conditions.
    • 💡Describe the practical sequence in order: spot, run, mark the front, locate spots, measure, calculate Rf, compare with standards.
    • 💡Name a suitable locating agent and say why it is needed, linking it to colourless spots such as amino acids.
    • 💡Explain why the same solvent and conditions are needed for the sample and standards when identifying substances.
    • 💡State that Rf is calculated from the origin to the centre of the spot divided by the origin to the solvent front.
    Common Mistakes
    • Thinking hydrogen bonds are covalent bonds within a water molecule; correct this by stating they are attractions between molecules, not shared-electron bonds.
    • Saying water is non-polar or that oxygen is δ⁺; correct this by stating oxygen is δ⁻ and hydrogen is δ⁺.
    • Confusing high specific heat capacity with high latent heat of vaporisation; correct this by linking specific heat capacity to resisting temperature change and latent heat to energy needed for a change of state.
    • Claiming ice is denser than water; correct this by stating ice is less dense and floats, insulating the water below.
    • Reversing the reactions, saying condensation adds water; correct this by stating condensation releases water and hydrolysis uses water.
    • Thinking a polymer is a single large monomer; correct this by stating a polymer is many monomers joined together.
    • Forgetting that one water molecule is released or used per bond formed or broken; correct this by linking the number of water molecules to the number of bonds.
    • Assuming hydrolysis only occurs in digestion; correct this by noting it also breaks down polymers inside cells when needed.
    • Assuming all biological molecules contain nitrogen: correction — carbohydrates and triglycerides contain only carbon, hydrogen and oxygen.
    • Thinking phosphorus is a component of proteins: correction — phosphorus is characteristic of nucleic acids and phospholipids, not of amino acids.
    • Treating sulphur as universal in proteins: correction — sulphur occurs only in the amino acids that contain it, such as cysteine and methionine, so only some proteins contain it.
    • Counting the ring oxygen as a carbon: correction — glucose's six-membered ring has five carbons plus one oxygen, and the sixth carbon lies outside the ring.
    • Assuming ribose is a six-carbon sugar because it is a nucleic acid component: correction — ribose is a pentose with five carbons forming a five-membered ring.
    • Thinking the alpha and beta forms differ in carbon number: correction — they differ only in the position of the –OH group on carbon 1 of the same six-carbon sugar.
    • Saying condensation releases carbon dioxide: correction — condensation of sugars releases water, not carbon dioxide.
    • Confusing which process adds water: correction — hydrolysis adds water and breaks bonds; condensation removes water and forms bonds.
    • Claiming a disaccharide contains one glycosidic bond per monosaccharide: correction — two monosaccharides are joined by a single glycosidic bond.
    • Describing cellulose as branched: correction — cellulose chains are straight and unbranched, held by hydrogen bonds.
    • Saying glycogen is made of β-glucose: correction — glycogen is a polymer of α-glucose.
    • Treating amylose and amylopectin as identical: correction — amylose is unbranched and helical, amylopectin is branched.
    • Treating starch and glycogen as identical: correct by noting glycogen is more highly branched than starch, giving faster glucose release.
    • Describing cellulose as branched or as made of α-glucose: correct by stating it is an unbranched polymer of β-glucose.
    • Saying glucose is insoluble so it can be stored: correct by noting glucose is soluble and is a transportable respiratory substrate, while storage polymers are insoluble.
    • Stating that a phospholipid has three fatty acids: correct by noting it has two fatty acids and one phosphate group.
    • Confusing ester bonds with peptide bonds: correct by identifying ester bonds between glycerol and fatty acids.
    • Saying condensation releases one water molecule per triglyceride: correct by noting three water molecules are released, one per bond formed.
    • Confusing condensation with hydrolysis: condensation releases water and builds the bond, whereas hydrolysis adds water and breaks it.
    • Thinking one water molecule is released per triglyceride: three ester bonds form, so three water molecules are released.
    • Placing the ester bond incorrectly, for example between two fatty acids: the ester bond links glycerol to each fatty acid.
    • Assuming breakdown requires no water: hydrolysis consumes water as each ester bond is split.
    • Describing a property without linking it to a function: always state what the property allows the molecule to do.
    • Claiming phospholipids are entirely hydrophobic: they are amphipathic, with a hydrophilic head and hydrophobic tails.
    • Saying cholesterol makes membranes rigid throughout: it regulates fluidity, reducing it at high temperatures and preventing stiffening at low temperatures.
    • Treating triglycerides as membrane components: their role is energy storage, insulation and protection, not bilayer formation.
    • Thinking the R group is the same in all amino acids: correct this by stating the R group is the variable side chain that distinguishes one amino acid from another.
    • Confusing the amine group with the carboxyl group: correct this by noting the amine group is —NH₂ (basic) and the carboxyl group is —COOH (acidic).
    • Placing the four groups on different carbon atoms: correct this by stating all four groups are bonded to the same central (alpha) carbon.
    • Saying condensation releases carbon dioxide: correct this by stating condensation releases water.
    • Thinking hydrolysis forms peptide bonds: correct this by stating hydrolysis breaks peptide bonds by adding water, while condensation forms them by removing water.
    • Believing the peptide bond forms between two R groups: correct this by stating it forms between the carboxyl carbon of one amino acid and the amine nitrogen of the next.
    • Confusing secondary and tertiary structure: secondary refers to local α-helix and β-pleated sheet folding held by backbone hydrogen bonds, whereas tertiary is the overall 3D fold of the whole chain held by R-group interactions.
    • Stating that quaternary structure applies to any folded protein: it requires two or more separate polypeptide chains associating, so a single-chain enzyme has no quaternary structure.
    • Claiming peptide bonds hold the tertiary structure: peptide bonds form the primary sequence, while tertiary folding depends on interactions between R groups.
    • Describing globular proteins as insoluble and fibrous: correction — globular proteins are compact and soluble because hydrophilic groups face the water.
    • Treating the prosthetic group as a separate protein: correction — it is a non-protein group attached to the polypeptide, such as the haem group in haemoglobin.
    • Saying shape is irrelevant to function: correction — the specific folded shape forms the active or binding site, so loss of shape (denaturation) loses function.
    • Describing fibrous proteins as compact and soluble: correct this by stating they are elongated and generally insoluble because of extensive cross-linking.
    • Confusing the roles of collagen and keratin: correct this by linking collagen's triple helix to tensile strength and keratin's disulfide bridges to hardness and waterproofing.
    • Claiming all fibrous proteins are identical in structure: correct this by distinguishing collagen's triple helix, keratin's cysteine cross-links and elastin's extensibility.
    • Confusing the role of sodium and potassium: Na⁺ is higher outside the cell and enters during depolarisation, while K⁺ is higher inside and leaves during repolarisation.
    • Stating that magnesium is part of haemoglobin; magnesium is in chlorophyll, whereas iron is in haemoglobin.
    • Saying that nitrate is used directly to make proteins; nitrate must first be reduced to ammonium before incorporation into amino acids.
    • Believing that hydroxide ions neutralise stomach acid in the small intestine; hydrogencarbonate is the main neutraliser, while OH⁻ contributes to alkaline conditions.
    • Forgetting the charge on phosphate; it is PO₄³⁻, not PO₄²⁻ or PO₄⁻, and this charge affects its binding in ATP and DNA.
    • Assuming Benedict's reagent detects all sugars directly: correction — it detects reducing sugars only; non-reducing sugars need acid hydrolysis and neutralisation first.
    • Forgetting to neutralise the acid before adding Benedict's reagent after hydrolysis: correction — sodium hydrogencarbonate must be added, or the acid prevents the colour change.
    • Confusing the iodine positive colour with the emulsion result: correction — iodine gives blue-black with starch; the emulsion test gives a cloudy white emulsion with lipid.
    • Treating a blue biuret result as positive: correction — blue is negative; purple indicates protein.
    • Confusing qualitative and quantitative tests: correction — a colour change alone is qualitative; quantitative methods measure how much is present.
    • Reading concentration directly from a single absorbance without a calibration curve: correction — the unknown must be compared with a curve built from standard solutions.
    • Ignoring controlled volumes and conditions: correction — consistent reagent volumes and conditions are needed for valid comparison.
    • Assuming one measurement is sufficient: correction — repeats and averaging reduce random error.
    • Thinking the solvent front is the top of the paper or plate: the solvent front is the furthest point the solvent itself has reached, and it must be marked before it dries.
    • Measuring Rf from the origin to the top of the paper rather than to the centre of the spot: the measurement is from the origin line to the centre of the spot.
    • Assuming Rf values are fixed constants for a substance: they depend on solvent, temperature and stationary phase, so conditions must be standardised for comparison.
    • Confusing which phase is stationary: in paper chromatography the cellulose holds water as the stationary phase, while the moving solvent is the mobile phase.
    • Spotting the sample below the solvent level: the sample must sit above the solvent so it is carried up the stationary phase rather than dissolving into the reservoir.
    • Using a pen to mark the origin or solvent front: pen ink runs in the solvent, so pencil must be used.
    • Measuring to the edge of a spot rather than its centre: measurement is to the centre of the spot for a reliable Rf.
    • Failing to run known standards alongside the sample: without standards under identical conditions, Rf values cannot reliably identify components.