AS & A Level Biology · Syllabus 9700 · Section 2 of 19

Biological molecules

Carbohydrates, lipids and proteins are built from small monomers joined by covalent bonds, and their functions follow from their shapes. This topic covers the bench tests for sugars, starch, lipids and proteins, the condensation and hydrolysis reactions that build and break polymers, the structure of starch, glycogen, cellulose, triglycerides, phospholipids and proteins, and the role of water.

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Syllabus coverage

  • 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
  • Describe and carry out a semi-quantitative Benedict's test on a reducing sugar solution by standardising the test and using the time to first colour change or a comparison with colour standards to estimate the concentration
  • Describe and carry out a test to identify the presence of non-reducing sugars, using acid hydrolysis and Benedict's solution
  • Describe and draw the ring forms of alpha-glucose and beta-glucose, and define the terms monomer, polymer, macromolecule, monosaccharide, disaccharide and polysaccharide
  • State the role of covalent bonds in joining smaller molecules together to form polymers, and describe the formation of a glycosidic bond by condensation and the breakage of a glycosidic bond by hydrolysis, with reference to disaccharides and the non-reducing sugar test
  • State that glucose, fructose and maltose are reducing sugars and that sucrose is a non-reducing sugar
  • Describe the molecular structure of the polysaccharides starch (amylose and amylopectin) and glycogen and relate their structures to their functions in living organisms
  • Describe the molecular structure of the polysaccharide cellulose and outline how the arrangement of cellulose molecules contributes to the function of plant cell walls
  • 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, and relate the molecular structure of triglycerides to their functions in living organisms
  • Describe the molecular structure of phospholipids with reference to their hydrophilic (polar) phosphate heads and hydrophobic (non-polar) fatty acid tails
  • Describe and draw the general structure of an amino acid and the formation and breakage of a peptide bond, explain the meaning of primary, secondary, tertiary and quaternary structure of proteins, and describe the hydrophobic interactions, hydrogen bonding, ionic bonding and covalent bonding including disulfide bonds that hold protein molecules in shape
  • State that globular proteins are generally soluble and have physiological roles and that fibrous proteins are generally insoluble and have structural roles, describe the structure of haemoglobin and of collagen and collagen fibres, and relate both structures to their functions
  • 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

Part of the AS & A Level Biology syllabus 9700.

The four biological moleculesAll four are built from a small number of repeating units. Every bond shown here is made by a condensation reaction, which removes onemolecule of water.Amino acid (the monomer of a protein)CRHH2NCOOHR group - it differs in every amino acidcentral carbonamino group (-NH2)carboxyl group (-COOH)hydrogen atomTriglyceride (a lipid)COCOCOglycerolglycerol backbonethree fatty acid chains (hydrocarbon tails)ester bond (condensation)fatty acidalpha-glucose and beta-glucoseEach ring is made of five carbons and one oxygen.OC1CH2OHOHHalpha-glucoseOH on C1 points below the ringOC1CH2OHOHHbeta-glucoseOH on C1 points above the ringNucleotide (the monomer of DNA and RNA)The three parts are joined by condensation, so one water moleculeis lost at each bond.Psugarbasephosphate grouppentose sugarnitrogenous baseDNA: deoxyribose, bases A T G C. RNA: ribose, uracil replaces thymine.
The key molecular structures of this topic drawn side by side: the ring form of alpha-glucose with six carbons and the -OH on carbon 1 pointing down, two alpha-glucose molecules joined by a glycosidic bond with a molecule of water released, a triglyceride built from glycerol and three fatty acids joined by ester bonds, a phospholipid with a hydrophilic phosphate head and two hydrophobic fatty acid tails, and the general structure of an amino acid with its amino group, carboxyl group, hydrogen and R group.

Testing for biological molecules: the four standard tests

Every one of these tests works the same way: a reagent is added to a sample and a colour change shows that a particular chemical group is present. The Benedict's test detects reducing sugars. Add an equal volume of Benedict's solution (blue, because it contains copper(II) sulphate in alkaline solution) to the sample and heat it in a water bath at about 80 °C for several minutes. The blue colour passes through green, yellow and orange to a brick-red precipitate as the amount of reducing sugar rises; the precipitate is copper(I) oxide, formed because the sugar reduces the copper(II) ions.

The iodine test detects starch. Add two or three drops of iodine in potassium iodide solution to the sample at room temperature and a blue-black colour appears. The colour is not the colour of iodine itself: iodine molecules slip inside the coiled helix formed by amylose and the trapped iodine changes the way light is absorbed. No other carbohydrate gives this result with iodine.

The emulsion test detects lipids. Shake the sample with about 2 cm3 of ethanol so that any lipid dissolves in it, then pour the ethanol into water. A lipid is soluble in ethanol but not in water, so it comes out of solution as billions of tiny droplets, and a cloudy white emulsion is seen. The cloudiness can be graded from faint to milky to judge roughly how much lipid is present.

The biuret test detects proteins and depends on the peptide bonds that join amino acids together. Add dilute sodium hydroxide to make the solution alkaline, then add a few drops of dilute copper(II) sulphate solution. A purple (lilac) colour appears because the copper(II) ions form a coloured complex with the nitrogen atoms of the peptide bonds. A single amino acid solution also gives a faint colour, but only peptides and proteins give the strong purple.

TestReagent and methodPositive result
Benedict's testAdd an equal volume of Benedict's solution and heat in a water bath at about 80 °CBlue to green, yellow, orange, then a brick-red precipitate
Iodine testAdd 2-3 drops of iodine in potassium iodide solution at room temperatureBlue-black
Emulsion testShake with ethanol, then pour the ethanol into waterA cloudy white emulsion
Biuret testAdd dilute sodium hydroxide, then a few drops of dilute copper(II) sulphate solutionPurple (lilac)
Learn each test as three things: the reagent, whether it needs heating, and the exact colour. The method and the result are usually separate marks.
The four food testsEach test uses a reagent that changes colour, or forms a precipitate or an emulsion, when the food contains the molecule being tested for.The tests are qualitative.TestReagentMethodPositive result (and what it shows)Reducing sugarBenedict's testBenedict's solution (blue), addedin excessAdd to the sample and heat in a water bathat about 80 degrees C for five minutes.Blue, then green, yellow, orange andfinally a brick-red precipitate. Thecolour shows how much reducing sugar ispresent.Starchiodine test (a test for starch)Iodine in potassium iodidesolution (yellow-brown)Add two or three drops to the sample at roomtemperature. No heating is needed.Yellow-brown to blue-black. The colouris lost again on heating, so the sampleis not heated.ProteinBiuret testDilute potassium hydroxide, thencopper(II) sulfate solution (paleblue)Make the sample alkaline with potassiumhydroxide, then add a few drops ofcopper(II) sulfate and shake.Pale blue to purple (lilac). The colourcomes from a complex between the copperions and the peptide bonds.Lipidemulsion testEthanol (alcohol), then waterShake the sample with ethanol, then pour theliquid into water.A cloudy white emulsion forms and thewater turns milky.
The four standard bench tests for biological molecules: a row of test tubes showing the Benedict's test changing from blue through green and orange to a brick-red precipitate, the iodine test giving a blue-black colour with starch, the emulsion test giving a cloudy white layer, and the biuret test turning purple with protein.

The semi-quantitative Benedict's test and testing for non-reducing sugars

A plain Benedict's test only tells you whether a reducing sugar is there. To make it semi-quantitative, standardise the method: keep the volume of sample, the volume of Benedict's solution, the temperature of the water bath and the length of the heating time all the same, and change only the concentration of the reducing sugar. There are two ways to use the result. The first is the time to first colour change: the more reducing sugar there is, the more quickly the blue colour disappears, because the copper(II) ions are reduced faster. The second is to make a set of colour standards from glucose solutions of known concentration and match the unknown sample to the closest standard.

The method is only as good as its controls, and the colour judgement is subjective. Two people looking at the same tube can disagree about whether the colour is yellow or orange, so the ranking of a set of tubes is more reliable than a single reading. A colorimeter makes the test quantitative: the wavelength is set to the red part of the spectrum because a reduced solution has lost its blue colour, and the higher the concentration of reducing sugar, the lower the absorbance (the higher the percentage transmission).

Sucrose gives a negative Benedict's result, so if a sample tests negative you must carry out a second test to find out whether a non-reducing sugar is present. Add dilute hydrochloric acid and heat the solution in a water bath: the acid hydrolyses the glycosidic bond, releasing glucose and fructose. Then neutralise the acid by adding sodium hydrogencarbonate until the solution stops fizzing and pH paper shows it is neutral. Finally add Benedict's solution and heat again. A brick-red precipitate in this second test shows that reducing sugar has been released, so the original solution contained a non-reducing sugar.

The neutralising step is the one candidates leave out. Benedict's solution works because its copper(II) ions are reduced in alkaline conditions, so if the acid is still in the tube the second test stays stubbornly blue and you would wrongly conclude that the sample contained no sugar at all. Never add both reagents at once either: the acid must do its work and then be neutralised, because the two steps do opposite things.

  • Standardise the test: same volume of sample, same volume of Benedict's solution, same water bath temperature, same heating time.
  • Time to first colour change: shorter time means more reducing sugar.
  • Colour standards: match the sample to a series of glucose solutions of known concentration.
  • Non-reducing sugar test: acid hydrolysis, then neutralise with sodium hydrogencarbonate, then Benedict's test.
  • Bench technique: label the tubes, use a separate sample for each test, and judge all the tubes in one sitting so the comparison is fair.
Acid first, then neutralise, then Benedict's. Skip the neutralisation and the acid keeps the solution acidic, so even a solution full of glucose stays blue.

Monosaccharides, disaccharides and the glycosidic bond

A monomer is a small molecule that can be joined to many other similar molecules. A polymer is a large molecule built from many monomers joined together, and a very large polymer is called a macromolecule. The monomers of the carbohydrates are the monosaccharides, single sugar units such as glucose and fructose with the formula C6H12O6. Two monosaccharides joined together form a disaccharide, and many joined together form a polysaccharide.

Glucose exists as a ring in solution, and it comes in two forms that you must be able to draw. In the ring, carbon 1 is joined to the oxygen that was on carbon 5, leaving an -OH group sticking out from carbon 1. In alpha-glucose that -OH group on carbon 1 points down; in beta-glucose it points up. Nothing else is different: the same six carbons, the same formula, the same everything except the position of one group. That single difference decides whether the polymer built from it coils into a helix or stays straight, and therefore whether the polymer is a storage molecule or a structural one.

Monomers are joined by condensation, in which a hydroxyl group of one sugar and a hydrogen of the other are removed together as a molecule of water, and a glycosidic bond forms between the two carbons. Glucose joined to glucose by an alpha-1,4 glycosidic bond gives maltose; glucose joined to fructose gives sucrose; glucose joined to galactose gives lactose. The bond is broken by hydrolysis, the addition of water, which is exactly what happens when amylase digests starch and when acid hydrolysis is used in the non-reducing sugar test.

The reducing sugars are the ones with a free aldehyde or ketone group that can reduce copper(II) ions. Glucose, fructose and maltose are all reducing sugars, which is why they turn Benedict's solution brick-red. Sucrose is a non-reducing sugar: the glycosidic bond joins carbon 1 of the glucose ring to carbon 2 of the fructose ring, so neither ring has a free reducing group left, and the copper(II) ions are never reduced.

DisaccharideMonomersBondReducing?
MaltoseGlucose + glucosealpha-1,4 glycosidicYes, reducing sugar
LactoseGlucose + galactosebeta-1,4 glycosidicYes, reducing sugar
SucroseGlucose + fructosealpha-1,2 glycosidicNo, non-reducing sugar
Draw alpha-glucose and beta-glucose identically, then change one thing: the -OH on carbon 1 points down in alpha, up in beta.

Starch, glycogen and cellulose: polymers of glucose with different jobs

Starch is the storage polysaccharide of plants and is a mixture of two polymers. Amylose is a long, unbranched chain of alpha-glucose joined by alpha-1,4 glycosidic bonds that coils into a helix. Amylopectin is also made of alpha-glucose, but along its chain there are alpha-1,6 glycosidic bonds that create branches. Starch is stored as grains in potato tubers, in the chloroplasts of leaves during the day, and in cereal grains such as rice, wheat and maize.

Every feature of starch suits it to storage. It is insoluble, so it does not dissolve in the cytoplasm, does not change the water potential of the cell and therefore does not draw water in by osmosis. It is a large molecule, so it cannot diffuse out of the cell and is not lost. It is compact, because the helical amylose packs a great deal of glucose into a small volume, so a plant can store a lot of energy without carrying a lot of bulk. The branches of amylopectin give many free ends, and because amylase and other hydrolytic enzymes work inwards from an end, the branching allows glucose to be released rapidly when the plant needs it for respiration.

Glycogen is the storage polysaccharide of animals and is the animal version of amylopectin, but with many more branches and shorter chains. The extra branching gives even more free ends, so glucose can be released even faster, which matters because animals have a much higher metabolic rate than plants and their muscles may need glucose within seconds. Glycogen is stored as granules in liver and muscle cells, and, like starch, it is insoluble and compact so it can be laid down without affecting the water potential of the cytoplasm.

Cellulose is the structural polysaccharide of plant cell walls and is built from beta-glucose. In the chain, every other beta-glucose molecule is turned through 180 degrees so that the -OH groups line up, and this makes the chain straight and unbranched instead of coiled. The chains are held side by side by hydrogen bonds between the -OH groups of neighbouring chains, and bundles of a few dozen chains form microfibrils. Long, straight, hydrogen-bonded microfibrils have a very high tensile strength, so the cell wall can resist the turgor pressure of the cell inside it and stop the cell bursting when it takes up water, and the criss-cross arrangement of the microfibrils keeps the cell's shape and supports the whole plant. The microfibrils are also spaced far enough apart for water and dissolved solutes to pass through the wall, so the wall is freely permeable.

PolysaccharideMonomer and bondsStructureRole
Amylose (starch)alpha-glucose, alpha-1,4 bondsUnbranched chain coiled into a helixCompact storage of glucose in plants
Amylopectin (starch)alpha-glucose, alpha-1,4 and alpha-1,6 bondsBranched chainStorage with many ends for rapid hydrolysis
Glycogenalpha-glucose, alpha-1,4 and alpha-1,6 bondsMore highly branched than amylopectin, shorter chainsRapid release of glucose in animals
Cellulosebeta-glucose, beta-1,4 bonds, alternate molecules invertedStraight unbranched chains held by hydrogen bonds into microfibrilsStrength and support in plant cell walls
Alpha and beta glucose differ only in the position of one -OH group, and that one difference is why starch is coiled, digestible and a good store, while cellulose is straight, hydrogen-bonded and strong.

Triglycerides and phospholipids: the non-polar molecules of life

A triglyceride is made of one molecule of glycerol joined to three fatty acids. Glycerol is an alcohol with three hydroxyl groups, and each fatty acid has a carboxyl group (-COOH) at one end and a long hydrocarbon tail at the other. Each fatty acid is joined to the glycerol by an ester bond, formed by a condensation reaction, so three molecules of water are released for every triglyceride made.

Fatty acids come in two kinds. A saturated fatty acid has no carbon to carbon double bonds in its hydrocarbon tail, so the tail is a straight chain and the molecules can pack closely together; fats built from them, such as butter, lard and the fat around a kidney, are solid at room temperature. An unsaturated fatty acid has one double bond (monounsaturated) or several (polyunsaturated), and each double bond puts a kink in the tail. The kinked molecules cannot pack closely, so the interactions between them are weaker and unsaturated oils, such as olive oil and sunflower oil, are liquid at room temperature.

Triglycerides are non-polar and hydrophobic. They have no charged groups and no groups carrying a partial charge that could form hydrogen bonds with water, so water molecules are not attracted to them and the two do not mix. This single property explains most of their functions. They are a compact energy store, because the long hydrocarbon tails are rich in carbon to hydrogen bonds and so release roughly twice as much energy per gram as carbohydrate when respired. They are insoluble, so stored fat does not affect the water potential of the cytoplasm and does not drag water into the cell by osmosis. They can also be stored without water, unlike glycogen, which holds water with it, and so save mass - one reason birds and desert animals rely on fat for migration and for survival.

Fat is also used for thermal insulation, as in the blubber of whales and seals and the subcutaneous fat of animals living in cold seas, and for protection, as in the fat that cushions the kidneys and the heart. Oxidising fat releases metabolic water, which is a real advantage for animals living in deserts, such as the camel and the kangaroo rat. A phospholipid is a modified triglyceride: glycerol is joined to two fatty acids and to a phosphate group. The phosphate group is charged, so the head is hydrophilic (polar) and mixes readily with water, while the two fatty acid tails are hydrophobic (non-polar). A molecule that has a water-loving end and a water-hating end arranges itself in water with the tails together and the heads facing the water, and this is the basis of the phospholipid bilayer of every cell membrane.

  • Triglyceride = glycerol + 3 fatty acids, joined by 3 ester bonds formed by condensation.
  • Saturated fatty acid: no C=C double bonds, straight tail, packs closely, higher melting point.
  • Unsaturated fatty acid: one or more C=C double bonds, kinked tail, packs loosely, lower melting point.
  • Triglyceride functions: energy store, insulation, protection and a source of metabolic water.
  • Phospholipid = glycerol + 2 fatty acids + a phosphate group, with a hydrophilic head and hydrophobic tails.
Non-polar means hydrophobic. With no charged groups to hydrogen bond with water, triglycerides do not dissolve - so they can be stored without changing the water potential of the cell.

Amino acids, peptide bonds and the four levels of protein structure

Every amino acid has the same general structure. A central carbon atom is bonded to an amino group (-NH2), a carboxyl group (-COOH), a hydrogen atom and an R group. The R group is the only part that differs from one amino acid to the next, and it is the R group that makes one amino acid acidic, another basic, another polar and another hydrophobic, or gives it a sulphur-containing group. In the conditions inside a cell the amino group carries a positive charge and the carboxyl group a negative charge, so you should draw them as -NH3+ and -COO-.

Two amino acids are joined by a condensation reaction between the carboxyl group of one and the amino group of the next. A molecule of water is released and a peptide bond forms between the carbon of the carboxyl group and the nitrogen of the amino group. Two amino acids joined this way make a dipeptide, and a chain of many makes a polypeptide. Peptide bonds are broken by hydrolysis: proteases hydrolyse them during digestion, and boiling with acid or alkali hydrolyses them too, which is why prolonged boiling destroys a protein permanently while ordinary heat treatment only denatures it.

Primary structure is the sequence and number of amino acids in the polypeptide chain. It is the most important of the four levels, because the R groups along the chain decide where every later interaction can form, so the primary structure determines the secondary, tertiary and quaternary structure. It is held by the strong covalent peptide bonds.

Secondary structure is the local folding of the chain into an alpha-helix or a beta-pleated sheet, held by hydrogen bonds between the partly negative oxygen of one C=O group and the partly positive hydrogen of an N-H group further along the same chain. Tertiary structure is the overall three-dimensional shape of the whole chain, held by four types of interaction between R groups: hydrogen bonds between polar R groups, ionic bonds between R groups carrying opposite charges, hydrophobic interactions where non-polar R groups are pushed together away from water, and covalent disulfide bonds between the sulphur atoms of two cysteine residues. Quaternary structure is the way two or more separate polypeptide chains, and any non-protein groups attached to them, are arranged together in one functional protein - as in haemoglobin, insulin and antibodies.

LevelWhat it isHeld together by
PrimaryThe sequence and number of amino acids in the chainPeptide bonds
SecondaryLocal coiling into an alpha-helix or folding into a beta-pleated sheetHydrogen bonds between C=O and N-H groups of the backbone
TertiaryThe overall 3D shape of one whole polypeptide chainHydrogen bonds, ionic bonds, hydrophobic interactions and disulfide bonds between R groups
QuaternaryThe arrangement of two or more polypeptide chains, plus any prosthetic groupsThe same interactions as the tertiary level, between and within the different chains
Heat, extremes of pH and heavy metal ions break the weak interactions of the tertiary structure and the protein is denatured. The peptide bonds of the primary structure survive all of them.

Haemoglobin and collagen: globular and fibrous proteins

Globular proteins are roughly spherical. The non-polar R groups are tucked away in the interior and the polar and charged R groups face outwards towards the water, so globular proteins are generally soluble. Their chains fold into a precise shape with a specific binding site, so they have physiological (metabolic) roles: enzymes, antibodies, hormones such as insulin and the oxygen carrier haemoglobin. Fibrous proteins are long chains lying parallel to one another and cross-linked at many points. They are generally insoluble and their chains form long, tough structures, so they have structural roles: collagen in tendons, bone and artery walls, keratin in hair, skin and nails, and elastin in the lungs. Solubility and role go together, and each follows from the amino acid sequence.

Haemoglobin is a globular protein with a quaternary structure built from four polypeptide chains: two alpha-globin chains and two beta-globin chains. Each chain is folded around a non-protein haem group, and at the centre of each haem group is an iron ion (Fe2+). The four haem groups give haemoglobin its dark red colour, and each iron ion can bind one molecule of oxygen, so one haemoglobin molecule can carry four molecules of oxygen. The importance of iron is that the oxygen binds reversibly to the Fe2+ ion: in the lungs, where the concentration of oxygen is high, oxygen joins to form oxyhaemoglobin, and in the tissues, where the concentration is low, the oxygen is released again. A person with iron deficiency anaemia has less iron available, so less haemoglobin is made and less oxygen can be carried, which is why they feel tired and short of breath on exertion. Because haemoglobin is globular and soluble, it can be packed into the cytoplasm of a red blood cell in very high concentration.

Collagen is the most abundant protein in a mammal and is a fibrous protein. Its primary structure is a repeating sequence of about a thousand amino acids in which every third amino acid is glycine, with a high proportion of proline and hydroxyproline in between. Glycine is the smallest amino acid and its single hydrogen atom can fit into the crowded centre of the molecule, which is why it must occur so regularly. Three of these chains wind round one another to form a triple helix, and the triple helices lie side by side and are cross-linked to each other by strong covalent bonds between their R groups, as well as by hydrogen bonds. The cross-linked bundles form collagen fibrils, and the fibrils are grouped into larger collagen fibres.

That structure explains every use of collagen. The covalent cross-links between the triple helices give collagen a very high tensile strength, so tendons, which join muscle to bone, can transmit the full force of a contracting muscle without stretching. The same strength, combined with a little flexibility, is why collagen is a major part of the matrix of bone, of cartilage at the joints and of the walls of arteries, which must stretch with every heartbeat and spring back without tearing. Collagen is insoluble, so it is not washed out of the tissue it strengthens, and its long fibres resist pulling but still allow the slight movement that a tendon, a joint or a blood vessel needs.

  • Globular: spherical, polar R groups outside, soluble, physiological role - haemoglobin, enzymes, antibodies.
  • Fibrous: long parallel chains, cross-linked, insoluble, structural role - collagen, keratin, elastin.
  • Haemoglobin: 2 alpha-globin + 2 beta-globin chains, 4 haem groups, 4 Fe2+ ions, so 4 oxygen molecules per molecule of haemoglobin.
  • Collagen: glycine every third residue, three chains wound into a triple helix, cross-linked into fibrils and then fibres.
Say it as a cause and a consequence: iron binds oxygen reversibly, so iron deficiency limits oxygen transport; covalent cross-links resist stretching, so collagen works in tendons and artery walls.

Water: hydrogen bonding and why life depends on it

In a water molecule the oxygen atom is far more electronegative than the two hydrogen atoms, so it pulls the shared electrons towards itself. The oxygen therefore carries a small partial negative charge and each hydrogen carries a small partial positive charge, and the molecule as a whole is polar. A hydrogen bond is the weak attraction between the partially positive hydrogen of one water molecule and the partially negative oxygen of another. Each water molecule can form hydrogen bonds with up to four others, so liquid water is not a collection of separate molecules but one enormous, constantly rearranging network.

Solvent action. Because water is polar, it is attracted to other polar and charged particles. Water molecules surround ions such as sodium and chloride and the polar groups of sugars and amino acids, and pull them apart, so they dissolve. This is why water is the solvent of life: almost every metabolic reaction in a cell happens in solution in the cytoplasm, blood plasma and cell sap, and dissolved substances can be transported from one part of an organism to another. Non-polar molecules such as lipids and the hydrocarbon parts of membranes do not dissolve in water, which is exactly why membranes form a barrier and why hydrophobic interactions hold a globular protein in shape.

High specific heat capacity. The specific heat capacity of water is about 4.2 J per gram per °C, which is much higher than that of most liquids. Raising the temperature of water means making its molecules move faster, but much of the heat energy supplied is used instead to break hydrogen bonds, so a large amount of energy produces only a small rise in temperature. Water therefore acts as a thermal buffer: the temperature of the cytoplasm stays stable even when the reactions inside the cell release heat; the temperature of a pond, lake or ocean changes only slowly as the air temperature changes, so aquatic organisms avoid lethal temperature swings; and a large body of water warms and cools the air over it, which is why coastal climates are mild. It also means that organisms themselves change temperature only slowly and can keep a steady body temperature.

Latent heat of vaporisation. The latent heat of vaporisation of water is about 2.26 kJ per gram, and it is high for the same reason: before a water molecule can escape from the surface as a vapour, the hydrogen bonds holding it to its neighbours must be broken. Because so much energy is needed to evaporate a small mass of water, evaporating water removes a great deal of heat from the surface it leaves. Sweating in mammals and panting in dogs use this to cool the body, and transpiration from leaves cools a plant and prevents in some part the leaves overheating in bright sunshine. It is also why water does not evaporate away from a pond as quickly as a non-hydrogen-bonded liquid would, which matters because a pond that dried out quickly would kill everything in it.

  • Water is polar: partial negative oxygen, partial positive hydrogens.
  • A hydrogen bond joins the partial positive hydrogen of one molecule to the partial negative oxygen of another; each molecule can form up to four.
  • Solvent action: polar and charged particles dissolve in water, so metabolic reactions take place in solution and substances can be transported.
  • High specific heat capacity: much of the heat is used to break hydrogen bonds, so water resists changes in temperature.
  • High latent heat of vaporisation: hydrogen bonds must be broken for molecules to escape, so evaporating water carries away a lot of heat.
All three properties come from one fact: water molecules form hydrogen bonds with each other. State the hydrogen bonding first, then the property, then the biological advantage.

Key terms you must be able to define

Monomer
A small molecule that can be joined to many other similar molecules to form a polymer, for example glucose, an amino acid or a fatty acid.
Polymer
A large molecule made of many monomers joined together by covalent bonds; a very large polymer, such as starch or a protein, is called a macromolecule.
Condensation reaction
A reaction that joins two molecules together and releases a molecule of water; it forms glycosidic bonds between sugars, ester bonds in triglycerides and peptide bonds in proteins.
Hydrolysis
A reaction that breaks a bond by adding water, used to break glycosidic bonds in digestion and in the acid hydrolysis step of the non-reducing sugar test.
Glycosidic bond
The covalent bond formed by condensation between the hydroxyl groups of two monosaccharides; alpha-1,4 and alpha-1,6 bonds occur in starch and glycogen, and beta-1,4 bonds in cellulose.
Reducing sugar
A sugar with a free aldehyde or ketone group that can reduce copper(II) ions in Benedict's solution to brick-red copper(I) oxide; glucose, fructose, maltose and lactose are reducing sugars, while sucrose is not.
Ester bond
The covalent bond formed by condensation between the carboxyl group of a fatty acid and a hydroxyl group of glycerol, so a triglyceride contains three ester bonds.
Peptide bond
The covalent bond formed by a condensation reaction between the carboxyl group of one amino acid and the amino group of the next, forming the primary structure of a polypeptide.
Disulfide bond
A strong covalent bond formed between the sulphur atoms of two cysteine residues, which holds parts of a protein's tertiary structure together and is not broken by heating.

Applied to the real world

Cambridge sets 9700 in unfamiliar contexts - a hospital, a research laboratory, a farm, a nature reserve. Use these as models for how to apply the biology, then write your own.

In a laboratory

A technician compares the reducing sugar content of several fruit juices using a semi-quantitative Benedict's test. The same volume of each juice and the same volume of Benedict's solution are placed in identical tubes, all of which are heated in one water bath at 80 °C, and a set of glucose standards of known concentration is run at the same time. The tubes are ranked by the time taken for the first colour change and then matched to the closest standard, so the juices can be placed in order from lowest to highest reducing sugar content rather than merely being shown to contain some sugar.

In a hospital

Blood plasma protein is measured in a hospital laboratory by the biuret test, because the purple colour is proportional to the number of peptide bonds present. A patient with nephrotic syndrome loses large amounts of the plasma protein albumin in the urine, and the biuret result falls well below the normal range. Someone with iron deficiency anaemia has a different problem: iron is needed to make the haem group of haemoglobin, so less haemoglobin is made, and although the plasma protein result is normal, less oxygen can be carried and the patient becomes tired and breathless on exertion.

In a food factory

Margarine is made from vegetable oils that are already rich in unsaturated fatty acids. The oil is hydrogenated by bubbling hydrogen through it at high temperature with a nickel catalyst, which removes C=C double bonds and converts unsaturated fatty acids into saturated ones. The kinks disappear from the hydrocarbon tails, the molecules pack together much more closely, and the liquid oil becomes a solid fat that spreads like butter but keeps for far longer. Food scientists count the change as a cost as well as a benefit, because saturated fat in the diet raises blood cholesterol more than unsaturated fat does.

On a farm

A dairy cow grazing on grass takes in a great deal of cellulose but produces no cellulase of its own. The rumen of the cow is a fermentation chamber containing dense populations of bacteria and protoctists that do produce cellulase, and they hydrolyse the beta-1,4 glycosidic bonds into glucose that the microorganisms respire. The cow then digests the microorganisms themselves, so the herd survives on the most abundant carbohydrate on land through a partnership of animal and microorganism. Silage and hay keep their cellulose content when they are stored, which is why they can be fed all winter.

In a tropical forest

Cellulose makes up the bulk of the wood in the canopy, because the microfibrils in each cell wall resist the turgor pressure of the cell within them and tube after tube of such cells builds the trunk. Water plays the other half of the story: a broadleaf canopy loses hundreds of litres a day through the stomata, and because water has a very high latent heat of vaporisation, each gram that evaporates removes a great deal of heat from the leaf. A tree whose leaves are in direct tropical sunlight is therefore cooled by the water it transpires, and its temperature stays within a range in which its enzymes still work.

In a nature reserve

A shallow pond in a reserve does not freeze or overheat as fast as the surrounding air. Water has a high specific heat capacity, so a large amount of energy must be lost or gained before the temperature of the pond changes much, and the pond therefore buffers the temperature of everything living in it. Insects, amphibians and fish in the pond avoid lethal temperature swings that would kill them in a smaller volume of water, and the same property means a wetland gradually warms and cools the air above it.

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Q1Give the reagent, the conditions and the positive result for the Benedict's test.
Add an equal volume of Benedict's solution and heat in a water bath at about 80 °C. A colour change from blue through green, yellow and orange to a brick-red precipitate shows a reducing sugar.
Q2Why must the hydrochloric acid be neutralised in the test for a non-reducing sugar?
Because Benedict's solution only works in alkaline conditions - its copper(II) ions cannot be reduced in acid. Without neutralisation the second test stays blue and a non-reducing sugar would be missed.
Q3Describe the emulsion test and its positive result.
Shake the sample with ethanol, then pour the ethanol into water. A cloudy white emulsion shows that a lipid is present, because lipid dissolves in ethanol but not in water.
Q4What is the only structural difference between alpha-glucose and beta-glucose?
The position of the -OH group on carbon 1: it points down in alpha-glucose and up in beta-glucose.
Q5What is a reducing sugar, and which sugars in this topic are reducing?
A sugar with a free aldehyde or ketone group that can reduce copper(II) ions to brick-red copper(I) oxide. Glucose, fructose, maltose and lactose are reducing; sucrose is not.
Q6How is a glycosidic bond formed and how is it broken?
Formed by condensation, when two monosaccharides join and a molecule of water is released. Broken by hydrolysis, when water is added - by amylase in digestion, or by dilute acid in the non-reducing sugar test.
Q7Give three ways in which the structure of starch suits it to storage.
It is insoluble, so it does not affect the water potential of the cell; it is a large molecule, so it cannot diffuse out; and its coiled, compact form stores a lot of glucose in a small volume. Its branches also give many ends for rapid hydrolysis.
Q8Why is glycogen more highly branched than amylopectin, and why does this matter?
Animal muscle and liver cells may need glucose within seconds. More branches mean more free ends, so hydrolytic enzymes can release glucose faster - which suits the higher metabolic rate of animals.
Q9Why can cellulose form strong fibres when starch cannot?
Cellulose is made of beta-glucose with alternate molecules inverted, so the chains are straight and unbranched, and hydrogen bonds form between the -OH groups of neighbouring chains. The resulting microfibrils have high tensile strength; coiled starch chains cannot pack in this way.
Q10How does a saturated fatty acid differ from an unsaturated one, and what does this do to melting point?
A saturated fatty acid has no C=C double bonds, so its tail is straight and the molecules pack closely, giving a higher melting point (solid fat). An unsaturated fatty acid has one or more C=C double bonds that kink the tail, so packing is poorer and the melting point is lower (liquid oil).
Q11How many fatty acids and phosphate groups does a phospholipid contain, and why does it form a bilayer?
Two fatty acids and one phosphate group, both joined to glycerol. The phosphate head is polar and hydrophilic while the two fatty acid tails are non-polar and hydrophobic, so in water the molecules line up with the tails inwards and the heads facing the water.
Q12Describe the quaternary structure of haemoglobin and state why iron is important.
Two alpha-globin chains and two beta-globin chains, each folded around a haem group containing an Fe2+ ion. Oxygen binds reversibly to the iron, so one haemoglobin molecule can carry four oxygen molecules and release them in the tissues.
Q13Which interactions hold the tertiary structure of a protein together?
Hydrogen bonds between polar R groups, ionic bonds between oppositely charged R groups, hydrophobic interactions between non-polar R groups, and covalent disulfide bonds between cysteine residues.
Q14Why does water have a high specific heat capacity and a high latent heat of vaporisation?
Water molecules form hydrogen bonds with one another. Much of any heat energy supplied is used to break those bonds rather than to speed the molecules up, so the temperature rises only slowly; and hydrogen bonds must be broken for a molecule to escape as a vapour, so evaporation needs a lot of energy.
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Exam-style questions

9 questions, 31 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

StateAO11 mark
State the colour of the precipitate formed when Benedict's solution is heated with a solution containing a reducing sugar.

Mark scheme — tick each point you actually made

Examiner tip: One mark, so one colour. Do not describe the method or the reagent here, and do not write 'blue to red' on its own: the blue is the starting colour of the reagent, not the result.
DefineAO12 marks
Define the terms monomer and polymer.

Mark scheme — tick each point you actually made

Examiner tip: Keep the two definitions in the same terms so that the relationship between them is clear. 'A polymer is a chain of monomers' is acceptable, but 'a polymer is a big molecule' is not.
OutlineAO23 marks
Outline how the structure of glycogen is related to its function as a storage molecule in a liver cell.

Mark scheme — tick each point you actually made

Any 3 of these score full marks. The examiner only needs 3 distinct points - pick the ones you actually made.
Examiner tip: Every point needs a feature and its consequence. 'Glycogen is branched' on its own scores nothing; 'branched, therefore many ends, therefore glucose is released rapidly' scores the mark.
ExplainAO24 marks
Explain how the molecular structure of cellulose makes it suitable for the function of the plant cell wall.

Mark scheme — tick each point you actually made

Any 4 of these score full marks. The examiner only needs 4 distinct points - pick the ones you actually made.
Examiner tip: Start from the monomer. The whole answer follows from beta-glucose and the inversion of alternate molecules, so name those first and then work outwards to the microfibril and the wall.
CalculateAO23 marks
A student completely hydrolysed 4.50 g of sucrose by heating it with dilute hydrochloric acid. Calculate the mass of glucose that could be produced. (Mr of sucrose = 342; Mr of glucose = 180.)

Mark scheme — tick each point you actually made

Examiner tip: Show every step. Use the Mr of glucose, not the Mr of sucrose, for the final multiplication, and check the size of the answer: half the mass of sucrose is a sensible result because one molecule of sucrose gives one molecule each of glucose and fructose.
ExplainAO24 marks
A solution of sucrose gives a negative result with Benedict's solution. Explain how you would show that this solution contains a non-reducing sugar.

Mark scheme — tick each point you actually made

Any 4 of these score full marks. The examiner only needs 4 distinct points - pick the ones you actually made.
Examiner tip: The order of the three steps is worth marks on its own: acid hydrolysis, then neutralisation, then Benedict's test. Say what the neutralisation is for - it is the reagent's requirement for alkaline conditions, not a way of stopping the reaction.
DescribeAO36 marks
A student is given a solution that may contain glucose, sucrose, starch, protein or lipid. Describe the tests the student should carry out, and the results expected, to find out which of these substances are present.

Mark scheme — tick each point you actually made

Any 6 of these score full marks. The examiner only needs 6 distinct points - pick the ones you actually made.
Examiner tip: This is a plan question, so describe what you do and what you see, in order. The mark for using fresh samples is easy to forget and it is the one that separates a bench answer from a written one.
DescribeAO24 marks
Describe the structure of a molecule of haemoglobin and explain how its structure is related to its function in the transport of oxygen.

Mark scheme — tick each point you actually made

Any 4 of these score full marks. The examiner only needs 4 distinct points - pick the ones you actually made.
Examiner tip: Name the two types of chain, count the chains and the haem groups, then link the iron to the reversible binding of oxygen. Saying haemoglobin 'carries four oxygens' without mentioning the four iron-containing haem groups loses a mark.
SuggestAO24 marks
Sweating cools a mammal and transpiration cools a leaf. Suggest, with reference to the properties of water, why the loss of only a small volume of water removes a large amount of heat.

Mark scheme — tick each point you actually made

Any 4 of these score full marks. The examiner only needs 4 distinct points - pick the ones you actually made.
Examiner tip: Start from hydrogen bonding every time. The chain is: hydrogen bonds, therefore a high latent heat of vaporisation, therefore a lot of heat removed, therefore cooling.

Exam technique for this topic

  • Give the method and the result separately in any bench test. 'Benedict's solution goes brick-red when heated' scores two things; 'you test it for reducing sugar' scores nothing.
  • Call the result of the Benedict's test a precipitate, not just a colour change. Copper(II) ions are reduced to copper(I) oxide, which is the brick-red solid that settles in the tube, and naming both the colour and the precipitate is often two marks.
  • In the non-reducing sugar test, write the three steps in the right order: hydrolyse with dilute hydrochloric acid, neutralise with sodium hydrogencarbonate, then carry out the Benedict's test. Forgetting the neutralisation is the single most common way to lose the explanation mark.
  • In a structure and function question, never state a feature on its own. Write the feature and its consequence in one sentence: highly branched, therefore many ends, therefore glucose is released rapidly. Two or three of these linked statements fill a four-mark answer.
  • Draw alpha-glucose and beta-glucose the same except for the -OH on carbon 1, and label it: down for alpha, up for beta. A drawing with no label on carbon 1 cannot score the comparison.
  • For water, water properties and biological roles are three separate steps. Name the hydrogen bonding, name the property it produces, and then name the biological advantage - a statement that 'water has a high specific heat capacity so organisms can survive' scores nothing without the link that a large energy change causes only a small temperature change.

Why students lose marks here

A student wroteA student wrote: 'Sucrose gives a negative result with Benedict's solution, so sucrose is not a sugar.'
Why it loses marksThe Benedict's test does not detect carbohydrates in general; it detects sugars that have a free reducing group. Sucrose contains two monosaccharides and is very much a sugar - it simply cannot reduce copper(II) ions.
The answer that scoresSucrose is a disaccharide of glucose and fructose joined by a glycosidic bond between carbon 1 of the glucose and carbon 2 of the fructose, so neither ring keeps a free reducing group. It is therefore a non-reducing sugar and gives a negative Benedict's result. Heating it with dilute hydrochloric acid hydrolyses the glycosidic bond, and after the acid has been neutralised with sodium hydrogencarbonate a Benedict's test on the products gives a brick-red precipitate, which shows that reducing sugars have been released.
A student wroteA student wrote: 'To test a solution for a non-reducing sugar, add dilute hydrochloric acid, then add Benedict's solution and heat the mixture.'
Why it loses marksThe acid is needed to hydrolyse the glycosidic bond, but it must be neutralised before the Benedict's test. Benedict's solution relies on copper(II) ions being reduced in alkaline conditions, and in acid conditions the test stays blue whatever the sugar content, so the student would report a false negative.
The answer that scoresAdd dilute hydrochloric acid and heat the solution in a water bath to hydrolyse the glycosidic bonds and release the monosaccharides. Then add sodium hydrogencarbonate a little at a time until the solution stops fizzing and pH paper shows it is no longer acidic. Only then add Benedict's solution and heat in a water bath: a brick-red precipitate in this second test shows that reducing sugar has been released, so the original solution contained a non-reducing sugar.
A laboratory bench set up for the semi-quantitative Benedict's test: a rack of labelled test tubes containing Benedict's solution and sugar solutions of different concentrations, a water bath held at 80 °C, a timer, a bottle of dilute hydrochloric acid and a bottle of sodium hydrogencarbonate for the non-reducing sugar test.
A student wroteA student wrote: 'Starch and cellulose are both polymers of glucose, so they have the same structure and both are digested by amylase.'
Why it loses marksStarch is built from alpha-glucose and cellulose from beta-glucose. The difference in the position of one -OH group makes alternate beta-glucose molecules in the chain turn through 180 degrees, so cellulose chains are straight and hydrogen bonded into microfibrils, while starch chains coil and branch. Amylase has an active site complementary to alpha-1,4 bonds and has no effect on beta-1,4 bonds.
The answer that scoresStarch is made of alpha-glucose joined by alpha-1,4 bonds, with alpha-1,6 branches in amylopectin, so the coils are compact and the branched ends allow rapid hydrolysis by amylase - good for storage. Cellulose is made of beta-glucose joined by beta-1,4 bonds, with alternate molecules inverted, so the chains are straight and are held in bundles by hydrogen bonds between their -OH groups. Those microfibrils have a high tensile strength and are resistant to hydrolysis, which is exactly what a plant cell wall needs. Humans digest neither beta-1,4 bonds nor the cellulose they form, because they have no cellulase.
A student wroteA student wrote: 'Boiling a protein solution denatures the protein because the heat breaks the peptide bonds in the chain.'
Why it loses marksHeat breaks the weak interactions that hold the folded shape - hydrogen bonds, ionic bonds and hydrophobic interactions - not the covalent peptide bonds. Peptide bonds are only broken by hydrolysis, which needs a protease or prolonged boiling with acid or alkali.
The answer that scoresThe primary structure is held by peptide bonds, which are strong covalent bonds and survive boiling, so the sequence of amino acids is unchanged. What heat destroys is the tertiary structure: the hydrogen bonds, ionic bonds and hydrophobic interactions between R groups break, so the chain unfolds and the precise shape of the molecule is lost. Any protein that depends on that shape, such as an enzyme with an active site or an antibody with a binding site, stops working. Disulfide bonds are also covalent and usually survive, which is why the primary structure can be recovered while the function cannot.

Where this fits in AS & A Level Biology

This page is one topic of Cambridge AS & A Level Biology 9700. AS & A Level Biology is the course home: every area and every topic in Cambridge's own order, the papers you sit and what each one tests, and the exam-practice tools. The full topic list, with each topic linked to its revision page, is on the AS & A Level Biology syllabus page.