The molecules of food: carbohydrates, fats, proteins and the enzymes that break them
By the end of this lesson you will be able to
- State the elements and the building blocks of carbohydrates, fats and oils, and proteins, and classify nutrients as macronutrients or micronutrients.
- Describe condensation and hydrolysis, and write word and symbol equations for making and breaking sucrose, starch, a triglyceride and a dipeptide.
- Calculate molar masses of food molecules and the masses that react in condensation and hydrolysis.
- Explain the difference between saturated and unsaturated fats at the molecular level, and why enzymes work fastest at an optimum temperature.
- Outline the food tests for starch, reducing sugars, protein, fat and unsaturation, with the colour change and one safety point for each.
You met this before
- Digestion breaks big food molecules into small soluble ones the blood can carry: amylase turns starch into maltose, proteases turn proteins into amino acids, lipase turns fats into fatty acids and glycerol (Year 3 digestion).
- Enzymes are biological catalysts: they speed up a reaction without being used up, and each one works on one substrate.
- A covalent bond is a shared pair of electrons; carbon forms four bonds, oxygen two, nitrogen three, hydrogen one (Year 3 covalent bonding).
- Molar mass is the sum of the relative atomic masses in the formula, in g mol⁻¹: C 12.01, H 1.008, N 14.01, O 16.00 (Unit 2); amount n = m ÷ M.
- Plants make protein from nitrates taken up by their roots — every protein molecule contains nitrogen (Year 3 nitrogen cycle).
Calculate the molar mass of glucose, C₆H₁₂O₆, and the amount in moles in a 4.0 g sugar cube (treat it as glucose).
Reveal
M = 6 × 12.01 + 12 × 1.008 + 6 × 16.00
M = 72.06 + 12.10 + 96.00
M = 180.16 g mol⁻¹
n = m ÷ M
n = 4.0 g ÷ 180.16 g mol⁻¹
n = 0.022 mol (2 significant figures, like the 4.0 g).
Revise it in full: Year 3 · Body systems · Lesson 1 — Digestion: from a plate of pap to a molecule in the blood · Year 3 · Atomic structure and bonding · Lesson 4 — Covalent bonding · Year 3 · Environmental cycles · Lesson 3 — Nitrogen: from air to protein
The puzzle
Chew a piece of plain white bread, without swallowing, for two full minutes. Something strange happens: it starts to taste sweet. Nothing sweet was added. The bread is mostly starch, which has no taste at all, yet your mouth is making sugar from it while you wait.
Now think of a spoonful of butter and a spoonful of sunflower oil. Both are mostly fat — and pure fat carries about 37 kJ per gram whichever it is — yet one is solid on the table and the other pours. Which molecules are hiding in food, what is happening in your mouth, and why do two fats behave so differently?
Predict what the sweet molecule made from starch in your mouth is, what made it, and why the bread must be chewed for a while before you notice it.
Compare your answer
Starch is a long chain of glucose units. The enzyme salivary amylase cuts the chain into two-glucose pieces — maltose, a sugar your tongue can taste. Enzymes need time and contact: chewing mixes saliva through the bread, and only after a minute or two has enough maltose built up to be noticed.
The reaction that splits the chain uses water — it is a hydrolysis — and it is the first step of the digestion you studied in Year 3.
The science — structured notes
1.1 Food is chemistry: six classes of nutrient
Every food is a mixture of a handful of kinds of molecule. Three are needed in large amounts and supply energy and building material — the macronutrients: carbohydrates, fats and oils (lipids), and proteins. Two are needed in tiny amounts but are essential — the micronutrients: vitamins and minerals. Water and dietary fibre complete the list.
| Nutrient | Elements | Building block | Main job in the body | Southern African and world sources |
|---|---|---|---|---|
| Carbohydrate | C, H, O (H : O = 2 : 1) | Monosaccharides (glucose, fructose, galactose) | Energy (about 17 kJ g⁻¹) | Maize meal (pap), sorghum (bogobe), rice, cassava, bread, sugar |
| Fats and oils | C, H, O (very little O) | Glycerol + three fatty acids | Energy store (about 37 kJ g⁻¹), insulation, cell membranes | Cooking oil, butter, groundnuts, marula nuts, avocado |
| Protein | C, H, O, N (often S) | Amino acids (20 kinds) | Growth and repair, enzymes, hormones, antibodies | Beans, seswaa (beef), mopane worms, eggs, milk, lentils, tofu |
| Vitamins | Varied | — | Small, specific jobs (vitamin C: collagen, antioxidant) | Fruit, vegetables, morogo (wild greens) |
| Minerals | Ions of Ca, Fe, Na, K, I, Zn… | — | Bones, blood (iron), nerves (sodium, potassium) | Milk, meat, salt, leafy greens |
Carbohydrates, fats and proteins are mostly macromolecules — very large molecules built from many small units joined by covalent bonds. That single idea explains most of this unit: food is made of chains, and cooking, digestion and spoilage are what happens when chains are made, broken or reshaped.
“Vitamins are micronutrients because their molecules are much smaller than those of proteins.”
Reveal answer
False. “Micro” refers to the amount needed — milligrams or micrograms a day — not the size of the molecule. Vitamin B12 has a molar mass of about 1 355 g mol⁻¹, far larger than glucose (180 g mol⁻¹), yet the body needs only a few micrograms of it a day.
The May 2022 Integrated Sciences paper asked exactly this classification: carbohydrates, proteins and fats as macronutrients; vitamins and minerals as micronutrients.
1.2 Carbohydrates: sugars and the chains made from them
Monosaccharides are single sugar units. Glucose and fructose share the formula C₆H₁₂O₆ but their atoms are arranged differently (they are isomers). In water most glucose molecules close into a six-membered ring of five carbon atoms and one oxygen atom.
Disaccharides are two units joined: maltose (glucose + glucose), sucrose — table sugar — (glucose + fructose) and lactose, the sugar in milk (glucose + galactose).
Two units join by condensation: an –OH group on one molecule and an H atom from an –OH on the other leave as a molecule of water, and the two units are linked through an oxygen atom (a glycosidic link). The reverse, splitting the link by adding water, is hydrolysis:
C₆H₁₂O₆ + C₆H₁₂O₆ → C₁₂H₂₂O₁₁ + H₂O (condensation)
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ (hydrolysis — by an enzyme or by acid)
Polysaccharides are chains of hundreds to many thousands of glucose units. Starch (in maize, sorghum, potatoes, rice) is the plant energy store: amylose is an unbranched coil, amylopectin is branched. Glycogen is the animal store in liver and muscle, even more branched. Cellulose is the plant cell wall; its glucose units are linked in a different orientation, human enzymes cannot hydrolyse it, and so it passes through us as dietary fibre.
Each glucose added to a chain loses one water molecule, so the repeating unit is C₆H₁₀O₅ and starch can be written (C₆H₁₀O₅)ₙ.
Sucrose is C₁₂H₂₂O₁₁. Calculate its molar mass and show that it equals the mass of glucose plus fructose minus water.
M(sucrose) = 12 × 12.01 + 22 × 1.008 + 11 × 16.00
M(sucrose) = 144.12 + 22.18 + 176.00
M(sucrose) = 342.3 g mol⁻¹
Check: glucose + fructose − water
180.16 + 180.16 − 18.02 = 342.30 g mol⁻¹ — the same, because condensation removes exactly one H₂O.
A plate of cooked maize porridge contains 81.0 g of starch. Assume all of it is hydrolysed to glucose in the gut. Calculate the mass of glucose formed and explain why it is greater than 81.0 g.
Unit. M(C₆H₁₀O₅) = 6 × 12.01 + 10 × 1.008 + 5 × 16.00
M(C₆H₁₀O₅) = 162.14 g mol⁻¹
Amount of units. n = 81.0 g ÷ 162.14 g mol⁻¹
n = 0.4996 mol of glucose units
Each unit becomes one glucose. m = 0.4996 mol × 180.16 g mol⁻¹
m = 90.0 g of glucose
Why heavier. Every unit gains one H₂O in hydrolysis: 0.4996 mol × 18.02 g mol⁻¹ = 9.00 g of water is built into the glucose. Mass is conserved — the water came from the gut.
A can of cola contains 39 g of sucrose. Calculate the amount of sucrose in moles to three significant figures, then the mass of glucose released when all of it is hydrolysed.
Reveal answer
n = 39 g ÷ 342.3 g mol⁻¹
n = 0.11394… mol, which is 0.114 mol to 3 s.f.
Each sucrose gives one glucose (and one fructose): n(glucose) = 0.114 mol
m = 0.114 mol × 180.16 g mol⁻¹
m = 20.5 g of glucose (and the same mass of fructose).
The first part is November 2023 Chemistry Q2(b), worth 2 marks: 1 for the division, 1 for rounding to three significant figures.
1.3 Fats and oils: glycerol, three fatty acids, and the double bond that decides
A fat or oil molecule is a triglyceride: one molecule of glycerol (propane-1,2,3-triol, C₃H₈O₃, with three –OH groups) joined by condensation to three fatty acids, long hydrocarbon chains ending in a –COOH group. Each join is an ester link and releases one water molecule:
glycerol + 3 fatty acids → triglyceride + 3 H₂O
A saturated fatty acid has only C–C single bonds in its chain — every carbon holds as many hydrogen atoms as it can. Stearic acid, C₁₇H₃₅COOH, is saturated. An unsaturated fatty acid has one or more C=C double bonds: oleic acid (olive oil) has one and is monounsaturated; linoleic acid (sunflower oil) has two and is polyunsaturated.
Why does a double bond turn a solid into a liquid? Straight saturated chains lie side by side like pencils in a box, touching along their whole length, so the weak attractions between molecules add up to a strong total — a high melting point.
In natural oils the double bond is almost always cis, which puts a fixed bend of about 30° in the chain. Bent chains cannot line up, touch less, attract less, and melt below room temperature.
Testing for unsaturation. Shake the fat or oil (dissolved in a solvent) with orange bromine water: bromine adds across each C=C bond and the orange colour disappears. A saturated fat leaves it orange. The more double bonds, the more bromine is decolourised — the basis of the “iodine value” that food chemists use to measure how unsaturated an oil is.
Margarine was first made cheaply by hydrogenating vegetable oils — adding hydrogen across some of the C=C bonds with a nickel catalyst. Explain why this turns an oil into a spreadable solid.
Reveal answer
Adding H₂ across a C=C bond converts it into a C–C single bond, so the chain loses its kink and becomes straight like a saturated chain.
Straight chains pack closely, the attractions between molecules are greater, and the melting point rises above room temperature. Partial hydrogenation also made some trans fats, whose unhealthy effect on blood cholesterol led many countries to restrict them — a change you will meet again in Lesson 4.
1.4 Proteins: chains of amino acids folded into shapes
Every amino acid has the same backbone — an amine group (–NH₂), a central carbon atom with a hydrogen atom, and a carboxylic acid group (–COOH) — plus a side group, R, that differs between the 20 kinds used in living things. Glycine has R = H; alanine has R = CH₃; cysteine has a sulfur atom in its R group, which is why many proteins contain sulfur.
Amino acids join by condensation between the –COOH of one and the –NH₂ of the next, making a peptide bond (–CO–NH–) and releasing water. Two joined make a dipeptide; tens to thousands make a polypeptide. The order of amino acids — the primary structure — is set by the DNA code.
A long chain does not stay straight. Hydrogen bonds between backbone groups make coils and sheets, and attractions between side groups — hydrogen bonds, ionic attractions, and covalent S–S “bridges” between cysteines — fold the whole chain into one specific three-dimensional shape. Shape is function: haemoglobin carries oxygen, collagen makes tendons, and enzymes catalyse because of their shape.
Glycine (C₂H₅NO₂) and alanine (C₃H₇NO₂) join by condensation. Write the equation and calculate the molar mass of the dipeptide.
Equation. C₂H₅NO₂ + C₃H₇NO₂ → C₅H₁₀N₂O₃ + H₂O
Atom check. C: 2 + 3 on the left, 5 on the right.
H: 5 + 7 on the left, 10 + 2 on the right; N: 2 on each side.
O: 2 + 2 on the left, 3 + 1 on the right. Balanced.
M = 5 × 12.01 + 10 × 1.008 + 2 × 14.01 + 3 × 16.00
M = 60.05 + 10.08 + 28.02 + 48.00
M = 146.15 g mol⁻¹ (glycine 75.07 + alanine 89.10 − water 18.02).
A small protein hormone has a chain of 51 amino acids. How many water molecules were released when the chain was assembled, and how many are needed to hydrolyse it completely into amino acids?
Reveal answer
51 amino acids in one chain are held by 50 peptide bonds; each bond formed released one H₂O, so 50 water molecules were released.
Complete hydrolysis breaks all 50 bonds and uses 50 water molecules. (Insulin actually has 51 amino acids in two chains joined by S–S bridges, so it has 49 peptide bonds — a good reason to read the structure before counting.)
1.5 Enzymes: proteins that catalyse by shape
An enzyme is a protein catalyst. Its folded chain leaves a pocket, the active site, whose shape and chemical groups match one substrate (Emil Fischer’s “lock and key”, 1894; we now know the site also moulds slightly around the substrate, the induced fit).
The substrate binds to form an enzyme–substrate complex, bonds in it are strained or brought together, the reaction happens with a lower activation energy, and the products leave. The enzyme is unchanged and ready again.
| Enzyme | Where it works | Hydrolyses | Products | Best pH |
|---|---|---|---|---|
| Salivary and pancreatic amylase | Mouth; small intestine | Starch | Maltose | about 7 |
| Pepsin (a protease) | Stomach | Proteins | Shorter polypeptides | about 2 |
| Trypsin (a protease) | Small intestine | Polypeptides | Short peptides, amino acids | about 8 |
| Lipase | Small intestine (fat emulsified by bile) | Triglycerides | Fatty acids + glycerol | about 8 |
| Lactase | Small intestine; added to “lactose-free” milk | Lactose | Glucose + galactose | about 6–7 |
Temperature. Warming speeds up the molecules, so enzyme and substrate collide more often and more complexes form each second: the rate roughly doubles for every 10 °C rise. Above the optimum (about 37–40 °C for human enzymes) the extra vibration breaks the hydrogen and ionic bonds that hold the fold. The active site loses its shape, the substrate no longer fits, and the rate falls steeply. The enzyme is denatured — usually permanently.
pH does the same from the other side: extra H⁺ or OH⁻ changes the charges on side groups, breaks ionic attractions and distorts the site. That is why pepsin works in stomach acid while amylase stops there.
What if you repeated the chewing-bread puzzle after rinsing your mouth with a little lemon juice (pH about 2.5)? Predict and explain.
Reveal answer
The sweet taste would appear much later or not at all. Salivary amylase works best near pH 7; at pH 2.5 the extra H⁺ ions alter the charges on its side groups, the ionic attractions holding its shape break, and the active site no longer fits starch — the enzyme is denatured, so little maltose forms.
This is exactly what happens when swallowed food reaches the acid of the stomach: starch digestion stops until pancreatic amylase takes over in the small intestine.
1.6 Testing food for its molecules
Five quick chemical tests tell you which molecules a food contains. Grind or crush the food in a little water first, so that the molecules are in solution or suspension, and always run a control — the same test on pure water — so you know what “no change” looks like.
| Test for | Reagent and method | Positive result | Negative result | Safety |
|---|---|---|---|---|
| Starch | A few drops of iodine solution (iodine in potassium iodide) | Blue-black | Stays orange-brown | Iodine stains skin and irritates eyes: goggles |
| Reducing sugars (glucose, fructose, maltose, lactose) | Equal volume of Benedict’s solution; heat in a water bath at about 80 °C for 5 minutes | Blue → green → yellow → orange → brick-red precipitate (more sugar, further along) | Stays blue | Water bath from a kettle, not a flame; tubes point away from faces |
| Protein | Biuret test: sodium hydroxide solution then a few drops of dilute copper(II) sulfate (or one biuret reagent) | Lilac / purple | Stays blue | Sodium hydroxide is corrosive: goggles, wash spills at once |
| Fat or oil | Emulsion test: shake with 2 cm³ of ethanol, pour the liquid into water | Cloudy white emulsion | Stays clear | Ethanol is highly flammable: no flames in the room |
| C=C (unsaturated fat) | Shake with bromine water (or iodine solution) | Orange → colourless | Stays orange | Bromine toxic: dilute, fume cupboard, gloves |
Why the tests work. Iodine molecules slip inside the coil of amylose and the complex absorbs most visible light. Reducing sugars reduce blue copper(II) ions in Benedict’s solution to insoluble red-brown copper(I) oxide, Cu₂O. The biuret test detects the peptide bond itself: in alkali, copper(II) ions bond to the N atoms of neighbouring peptide links, giving a lilac complex — so it works for any protein but not for single amino acids.
Fats dissolve in ethanol but not in water; poured into water, the fat comes out as tiny droplets that scatter light.
A student writes: “I added Benedict’s solution to the crushed maize and it stayed blue, so maize contains no carbohydrate.” Identify two errors in the reasoning.
Reveal answer
First, Benedict’s solution detects only reducing sugars; maize’s carbohydrate is mostly starch, which needs the iodine test.
Second, the test was not heated: Benedict’s solution must be warmed in a water bath at about 80 °C for five minutes before any colour change can be judged. “Stayed blue” without heating proves nothing.
Olive oil is mostly triolein, the triglyceride of oleic acid, C₅₇H₁₀₄O₆, which has three double bonds per molecule. Calculate the mass of bromine, Br₂, that 1.00 g of pure triolein can decolourise, and the “iodine value” (the mass of I₂, in g, that adds to 100 g of fat).
Relative atomic masses: Br 79.90; I 126.90.
Reveal answer
M(triolein) = 57 × 12.01 + 104 × 1.008 + 6 × 16.00
M(triolein) = 684.57 + 104.83 + 96.00
M(triolein) = 885.40 g mol⁻¹
n(triolein) = 1.00 g ÷ 885.40 g mol⁻¹
n(triolein) = 1.129 × 10⁻³ mol
One Br₂ adds across each of the three double bonds.
n(Br₂) = 3 × 1.129 × 10⁻³ mol
n(Br₂) = 3.388 × 10⁻³ mol
m(Br₂) = 3.388 × 10⁻³ mol × 159.80 g mol⁻¹
m(Br₂) = 0.541 g
Iodine value: n(triolein) in 100 g = 100 g ÷ 885.40 g mol⁻¹ = 0.11294 mol
n(I₂) = 3 × 0.11294 mol = 0.3388 mol
m(I₂) = 0.3388 mol × 253.80 g mol⁻¹
m(I₂) = 86.0 g per 100 g — real olive oils give 75–94, because they also contain some saturated and some doubly unsaturated chains.
Try it — an enzyme and the temperature it prefers
Potato cells contain the enzyme catalase, which breaks down hydrogen peroxide, a harmful by-product of metabolism: 2H₂O₂ → 2H₂O + O₂. Pour hydrogen peroxide on potato and oxygen foams up; the height of foam after 60 s measures how fast the enzyme worked.
The lab lets you set the temperature from 10 to 70 °C, and keep the pH and the mass of potato as controls. It is the same experiment a food technologist would use to check that vegetables have been blanched enough.
Mean foam heights: 10 °C 9 mm; 20 °C 18 mm; 30 °C 37 mm; 40 °C 59 mm; 50 °C 39 mm; 60 °C 1 mm. State the optimum range and explain the 60 °C value in terms of the active site. A frozen-food factory blanches peas in water at 95 °C: what does this data suggest about the peas’ own enzymes after blanching?
Reveal answer
The optimum lies between 30 and 50 °C, close to 40 °C, where the height is greatest (59 mm). Each 10 °C rise from 10 to 30 °C doubles the rate (9 → 18 → 37 mm).
At 60 °C the foam is almost zero because the bonds holding the enzyme’s folded shape have broken: the active site no longer fits hydrogen peroxide, so almost no complexes form — denaturation.
At 95 °C the peas’ enzymes are denatured even faster, which is the point of blanching: the enzymes that would otherwise spoil colour and flavour in the freezer are destroyed.
Common mistakes to avoid
- “Carbohydrates contain carbon and water” — they contain the elements C, H and O; no water molecules are present in glucose.
- Forgetting the water in condensation and hydrolysis — every link made releases one H₂O; every link broken uses one.
- Calling all sugars “reducing” — sucrose is not; test starch with iodine, not Benedict’s.
- Saying the enzyme is “killed” or “dead” — say “denatured: the active site changes shape”.
- Claiming the C=C bond is weak — melting separates molecules; it is the kink, and the weaker forces between molecules, that lowers the melting point.
- Proteins contain only C, H and O — every protein contains nitrogen, and most contain sulfur.
- Benedict’s without heat — the result is meaningless unless the tube is heated in a water bath.
Study smart: how this is examined
Study routine. Today: make one card per macronutrient (elements, building block, link, test) and redo the two EXAMPLES without looking. In two days: write the condensation equations for sucrose, a triglyceride and a dipeptide from memory, and check the atoms balance.
In a week: run the catalase lab through all seven temperatures, then the Think task and the quiz.
How we found out — from digesting bread to crystals of enzyme
In 1833 two French chemists, Anselme Payen and Jean-François Persoz, extracted from germinating barley a substance that turned starch into sugar without being used up. They called it diastase — the first enzyme ever isolated, the amylase that makes chewed bread sweet.
But were such “ferments” chemicals, or did they need living cells? Louis Pasteur argued that fermentation needed life. In 1897 Eduard Buchner ground yeast with sand, pressed out the juice, and showed that the cell-free juice still fermented sugar to alcohol. The catalysts were molecules; he won the 1907 Nobel Prize in Chemistry.
Emil Fischer worked out the structures of glucose, fructose and their relatives in the 1880s and 1890s. In 1894 he noticed that enzymes act on one sugar but not its mirror-image twin, and proposed the lock and key. In 1902 he and Franz Hofmeister independently proposed that proteins are chains of amino acids joined by peptide bonds.
Many still doubted that enzymes were proteins until James Sumner crystallised the enzyme urease from jack beans in 1926 and showed the crystals were pure protein (Nobel Prize, 1946).
Protein itself had to be measured before it could be understood. In 1883 Johan Kjeldahl, at the Carlsberg brewery laboratory in Copenhagen, found a way to convert all the nitrogen in a sample into ammonia and measure it. Because food proteins are about 16 % nitrogen, protein ≈ nitrogen × 6.25 — a method still used worldwide.
Its weakness was exposed in 2008, when melamine, a nitrogen-rich industrial chemical, was added to diluted milk in China to fake a high protein reading. Hundreds of thousands of infants fell ill and several died; food laboratories now test for the protein, not just the nitrogen.
Where it is used now
- Lactose-free milk. Most adults worldwide, especially in East Asia and much of Africa, produce less lactase after childhood; dairies add lactase to hydrolyse lactose into glucose and galactose, which is why lactose-free milk tastes sweeter.
- Maize and sorghum. Malting sorghum for traditional beer and for porridge sets the grain’s own amylases working on its starch — the same chemistry as chewing bread, used across southern Africa for centuries.
- Nutrition labels. Every carbohydrate, fat and protein figure on a label comes from the chemistry in this lesson: fat extracted with a solvent, protein from Kjeldahl nitrogen, sugars by enzyme or chromatography.
- Oils and health. Food chemists grade oils by unsaturation (the iodine value) because it predicts melting point and how fast an oil goes rancid (Lesson 4).
- Detergents and bakeries. Proteases and lipases in “biological” washing powders digest food stains; amylases keep bread soft for longer.
Science now — and the scientist you need to be
What is happening in 2026
Protein structure, the puzzle Fischer and Sumner opened, has been transformed by computation. In 2020 DeepMind’s AlphaFold predicted the shapes of proteins from their amino acid sequences with accuracy close to experiment, and a database of predicted structures for over 200 million proteins was released in 2022. The 2024 Nobel Prize in Chemistry was shared by David Baker for designing new proteins and by Demis Hassabis and John Jumper for predicting structures.
Food scientists now use these tools to redesign enzymes that survive baking temperatures, to understand how plant proteins might be folded to feel like meat (Lesson 5), and to predict which proteins might trigger allergies. The chemistry underneath is still the one in this lesson: a chain, folded by weak bonds into a shape that does a job.
The mindshift
What you need to be able to do
Identify the building blocks of each macronutrient and write the condensation or hydrolysis equation with the water in it; calculate molar masses and amounts to the requested significant figures; explain melting points and enzyme activity at the level of molecules; and choose the right food test, with its control, colour and safety point.
See an example
A typical sunflower oil label: fat 100 g, of which saturates about 11 g. A typical biscuit: carbohydrate 65 g, of which sugars 22 g; fat 20 g, of which saturates 9 g; protein 7 g.
A good answer: coconut oil is about 80–90 % saturated fat, much of it lauric acid chains, which pack closely; sunflower oil is mostly unsaturated linoleic and oleic chains with cis kinks, which cannot pack. An answer that says “the bonds in sunflower oil are weaker” has confused covalent bonds with the forces between molecules.
Key terms
| Macronutrient | A nutrient needed in large amounts: carbohydrate, fat, protein. |
| Micronutrient | A nutrient needed in small amounts: vitamins and minerals. |
| Monosaccharide | A single sugar unit, such as glucose or fructose (C₆H₁₂O₆). |
| Polysaccharide | A long chain of sugar units: starch, glycogen, cellulose. |
| Condensation reaction | Two molecules join with the loss of a small molecule, usually water. |
| Hydrolysis | Breaking a bond by reaction with water; the reverse of condensation. |
| Triglyceride | A fat or oil molecule: glycerol joined to three fatty acids by ester links. |
| Saturated / unsaturated | A chain with only C–C single bonds / with at least one C=C double bond. |
| Peptide bond | The –CO–NH– link between two amino acids, formed by condensation. |
| Enzyme | A protein that catalyses a specific reaction by binding the substrate in its active site. |
| Denaturation | Permanent loss of a protein’s three-dimensional shape when the bonds holding it break. |
| Reducing sugar | A sugar that reduces blue Cu²⁺ in Benedict’s solution to red Cu₂O on heating. |
Think like a scientist
Criterion C A student uses Benedict’s solution and a colorimeter to measure the sugar released when amylase digests starch.
She mixes 5 cm³ of 1 % starch with 1 cm³ of amylase at 35 °C, takes a sample every 2 minutes, tests it with Benedict’s solution, and reads the percentage of light absorbed at a red filter.
She first makes a calibration line with known maltose solutions: absorption rises by 11.0 % for every 1.0 mg cm⁻³ of maltose, from 2.0 % at zero.
| Time / min | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| Absorption, trial 1 / % | 2.1 | 18.4 | 31.8 | 41.5 | 47.7 | 51.2 | 52.6 |
| Absorption, trial 2 / % | 1.9 | 17.8 | 32.4 | 40.9 | 47.1 | 50.6 | 52.8 |
| Absorption, trial 3 / % | 2.0 | 18.1 | 38.9 | 41.2 | 47.4 | 50.9 | 52.4 |
(a) Identify the anomalous result and suggest a specific cause. [2]
(b) Calculate the mean absorption at each time, excluding the anomaly, to one decimal place. [2]
(c) Convert the mean absorption at 4 min and at 12 min into maltose concentration using the calibration line. [2]
(d) Describe the trend, quoting values, and explain the shape of the curve in terms of substrate and active sites. [4]
(e) Calculate the mean rate of maltose production in the first 4 minutes and in the last 4 minutes, in mg cm⁻³ min⁻¹. [2]
(f) Explain one systematic error and one random error in this method, and one improvement for each. [4]
Model answer
(a) Trial 3 at 4 min, 38.9 % — about 7 % above 31.8 and 32.4 [1].
Cause: that sample was left longer in the Benedict’s water bath (or the bath was hotter), so more Cu₂O formed — specifically a timing or temperature difference in the test, not in the digestion [1].
(b) 0 min 2.0; 2 min 18.1; 4 min 32.1 (from 31.8 and 32.4 only); 6 min 41.2; 8 min 47.4; 10 min 50.9; 12 min 52.6 % [2; 1 if the anomaly is included].
(c) Concentration = (absorption − 2.0) ÷ 11.0
At 4 min: (32.1 − 2.0) ÷ 11.0 = 2.7 mg cm⁻³ [1]
At 12 min: (52.6 − 2.0) ÷ 11.0 = 4.6 mg cm⁻³ [1]
(d) The absorption rises quickly at first — by 16.1 % in the first 2 minutes — then more and more slowly, only 1.7 % between 10 and 12 min, approaching about 53 % [1].
At the start there is plenty of starch, so almost every active site is occupied and complexes form at the maximum rate [1].
As starch is hydrolysed its concentration falls, so collisions between substrate and active sites become less frequent and the rate falls [1].
The curve levels off as the starch is used up; the enzyme is not used up [1].
(e) First 4 min: maltose rises from 0.0 to 2.7 mg cm⁻³
Rate = 2.7 ÷ 4 = 0.68 mg cm⁻³ min⁻¹ [1]
Last 4 min: from (47.4 − 2.0) ÷ 11.0 = 4.1 to 4.6 mg cm⁻³
Rate = (4.6 − 4.1) ÷ 4 = 0.12 mg cm⁻³ min⁻¹ — about six times slower [1].
(f) Systematic: the enzyme keeps digesting starch in each sample until the Benedict’s solution is heated, so every sample reads a little later than its label — stop the reaction at once by adding each sample to the hot Benedict’s solution, or to acid [2].
Random: the colorimeter reading varies with scratches and fingerprints on the cuvette and with how well the Cu₂O precipitate is suspended — wipe and orient the cuvette the same way and shake it just before every reading [2]. (Vague answers such as “mistakes by the student” or “more trials” score 0.)
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth. Relative atomic masses: C 12.01, H 1.008, N 14.01, O 16.00.
State the elements present in all proteins and name the bond that joins their building blocks.
Reveal answer
Carbon, hydrogen, oxygen and nitrogen (most also contain sulfur) [1].
The peptide bond (amide link, –CO–NH–) [1].
Calculate the amount in moles of lactose, C₁₂H₂₂O₁₁, in 250 cm³ of milk containing 12 g of lactose. Give your answer to three significant figures.
Reveal answer
M = 342.3 g mol⁻¹ (same formula as sucrose)
n = 12 g ÷ 342.3 g mol⁻¹ [1]
n = 0.0351 mol (3.51 × 10⁻² mol) [1]
Explain why beef fat is solid at room temperature but sunflower oil is liquid.
Reveal answer
Beef fat has mostly saturated fatty-acid chains with only C–C single bonds, which are straight [1].
Sunflower oil has mostly unsaturated chains with cis C=C bonds that put kinks in the chains [1].
Straight chains pack closely with strong total forces between molecules, so more energy is needed to separate them — a higher melting point; kinked chains cannot pack, so the forces are weaker and the melting point is below room temperature [1].
Outline how you would show that a white powder contains protein but no starch, including the result expected and a control.
Reveal answer
Dissolve or suspend the powder in water in two tubes; to one add sodium hydroxide solution and then drops of copper(II) sulfate (biuret test): it turns lilac, so protein is present [1].
To the other add iodine solution: it stays orange-brown, so no starch [1].
Control: the same tests on distilled water alone, to show the colours without the powder (blue and orange-brown) [1].
Pepsin works in the stomach at pH 2 but not in the small intestine at pH 8. Explain why.
Reveal answer
The shape of pepsin’s active site depends on ionic and hydrogen bonds between side groups [1].
At pH 8 the concentration of H⁺ is a million times lower than at pH 2, so the charges on some side groups change and those bonds break or form differently [1].
The active site changes shape, protein no longer fits, and pepsin is denatured or inactive [1].
A triglyceride is made from glycerol (C₃H₈O₃) and three molecules of palmitic acid (C₁₆H₃₂O₂). Deduce its molecular formula.
Reveal answer
Add the atoms and subtract three water molecules [1].
C: 3 + 48 = 51
H: 8 + 96 − 6 = 98
O: 3 + 6 − 3 = 6
Formula: C₅₁H₉₈O₆ (tripalmitin) [1].
7. A 2.00 g sample of dried infant formula is analysed by the Kjeldahl method and found to contain 0.0512 g of nitrogen. Calculate the protein percentage using protein = N × 6.25. Then calculate the mass of melamine, C₃H₆N₆, that a fraudster would need to add to 2.00 g of formula containing only 5.0 % real protein to make the test read the same, and explain why this fraud worked.
Reveal answer
Protein = 0.0512 × 6.25 = 0.320 g
Percentage = 0.320 ÷ 2.00 × 100 = 16.0 % [1]
Genuine: 5.0 % of 2.00 g = 0.100 g of protein, containing 0.100 ÷ 6.25 = 0.0160 g of N
Nitrogen still needed = 0.0512 − 0.0160 = 0.0352 g [1]
M(melamine) = 3 × 12.01 + 6 × 1.008 + 6 × 14.01 = 126.14 g mol⁻¹
Fraction of nitrogen = 84.06 ÷ 126.14 = 0.666 (66.6 %)
Mass of melamine = 0.0352 ÷ 0.666 = 0.053 g [1]
The test measured total nitrogen, not protein: it assumed every nitrogen atom came from peptide chains. Melamine carries far more nitrogen per gram than protein (66.6 % against 16 %), so a small, cheap addition faked a large protein content [1].
8. Cellulose and starch are both polymers of glucose with repeating unit C₆H₁₀O₅. A cow can live on grass, but a human cannot. Explain, using the ideas of this lesson, why the same glucose units are food for one and fibre for the other, and calculate the mass of glucose a cow’s gut microbes could release from 1.00 kg of cellulose.
Reveal answer
In cellulose the glucose units are linked in a different orientation from starch, giving a different shape of chain [1].
An enzyme’s active site fits only one shape of substrate; human amylase fits the starch link but not the cellulose link, and humans make no cellulase — so cellulose passes through as fibre [1].
Microbes in the cow’s rumen make cellulase, whose active site fits the cellulose link, and hydrolyse it.
n(units) = 1 000 g ÷ 162.14 g mol⁻¹ = 6.168 mol [1]
m(glucose) = 6.168 mol × 180.16 g mol⁻¹ = 1.11 kg (1 111 g) [1]
Summary
- Carbohydrates (C, H, O) are sugars and chains of sugars; fats and oils (C, H, O) are glycerol plus three fatty acids; proteins (C, H, O, N, often S) are chains of amino acids.
- Condensation joins units and releases water; hydrolysis breaks links and uses water — the mass arithmetic always includes 18.02 g mol⁻¹ of H₂O per link.
- Saturated chains are straight and pack closely (solid fats); cis C=C bonds kink unsaturated chains so they cannot pack (liquid oils). Bromine water detects C=C.
- A protein’s function comes from its folded shape; enzymes are protein catalysts whose active site fits one substrate, with an optimum temperature and pH and permanent denaturation beyond it.
- Food tests: iodine (starch, blue-black), Benedict’s heated (reducing sugars, brick-red), biuret (protein, lilac), emulsion (fat, cloudy white) — always with a water control.
Next time → Energy in food: why a gram of fat carries more than twice the energy of a gram of sugar, how to measure it by burning a peanut under a can of water, and how to read the kilojoules on a food label.