AQA A-level Biology: Monomers, Polymers and Carbohydrates (7402)

This guide teaches sections 3.1.1 (Monomers and polymers) and 3.1.2 (Carbohydrates) of the AQA AS and A-level Biology specification, version 1.5, for exams from June 2016 onwards. Sections 1 to 4 of the specification are the AS content and are also the first year of the A-level, so this material is examined in both. Work through it with the AQA A-level Biology hub and the printable checklist.

When you have read it, condense it with the revision notes and test yourself with the practice questions. The whole of section 3.1 is summarised in the biological molecules study guide, and the free diagnostics show which topics need work first.

Outcomes covered

Section	Outcome
3.1.1	Define monomer and polymer; give monosaccharides, amino acids and nucleotides as examples; define condensation and hydrolysis
3.1.2	Name glucose, galactose and fructose; describe the glycosidic bond; give the make-up of maltose, sucrose and lactose
3.1.2	Describe alpha-glucose and beta-glucose, and which polysaccharides each forms
3.1.2	Describe the structure and function of glycogen, starch and cellulose, and relate structure to function in animal and plant cells
3.1.2	Use Benedict’s solution (reducing and non-reducing sugars) and iodine in potassium iodide (starch)

3.1.1 Monomers and polymers

A monomer is a smaller unit from which larger molecules are made. A polymer is a molecule made from a large number of monomers joined together. The specification names three examples of monomer.

Monomer	Polymer it makes	Where you meet it
Monosaccharide	Polysaccharide	Section 3.1.2 (this guide)
Amino acid	Polypeptide, then protein	Section 3.1.4
Nucleotide	Polynucleotide (DNA, RNA)	Section 3.1.5

The chemistry is the same each time, which is why the topic is worth learning properly.

• A condensation reaction joins two molecules together with the formation of a chemical bond. It involves the elimination of a molecule of water.
• A hydrolysis reaction breaks a chemical bond between two molecules. It involves the use of a water molecule.

The two reactions are opposites. Building a polymer is a series of condensations. Digesting or breaking it down is a series of hydrolyses. In a cell, enzymes catalyse both.

Counting bonds and water molecules

Joining n monomers needs n − 1 bonds, so n − 1 molecules of water are released. Breaking the polymer back to n monomers uses n − 1 molecules of water. This holds even if the polymer is branched: each new bond joins one more unit to the structure.

Worked example 1. A chain of 25 glucose molecules is built by condensation. Glucose is C₆H₁₂O₆. Find the water released and the molecular formula of the chain.

1. Bonds formed = 25 − 1 = 24, so 24 water molecules are released.
2. Carbon: 25 × 6 = 150. Hydrogen: (25 × 12) − (24 × 2) = 300 − 48 = 252. Oxygen: (25 × 6) − 24 = 126.
3. The chain is C₁₅₀H₂₅₂O₁₂₆.
4. Check with relative molecular mass, taking glucose as 180 and water as 18: (25 × 180) − (24 × 18) = 4500 − 432 = 4068.

3.1.2 Monosaccharides

Monosaccharides are the monomers from which larger carbohydrates are made. Glucose, galactose and fructose are common monosaccharides. All three have the molecular formula C₆H₁₂O₆, but their atoms are arranged differently, so they are different sugars.

Alpha-glucose and beta-glucose

Glucose has two isomers, α-glucose and β-glucose. Isomers have the same molecular formula but different structures. In solution glucose forms a ring of five carbon atoms and one oxygen atom. Carbon 6 sits outside the ring as a CH₂OH group. Only carbon 1 distinguishes the two forms.

Position	α-glucose	β-glucose
OH on carbon 1	Sits under the ring	Sits over the ring
H on carbon 1	Sits over the ring	Sits under the ring
Carbons 2 to 6	Identical	Identical

Exam answers lose marks for “the OH is in a different place”. Name carbon 1 and say above or below the ring.

Disaccharides and the glycosidic bond

A condensation reaction between two monosaccharides forms a glycosidic bond, a C–O–C link, and releases one water molecule. The product is a disaccharide.

Disaccharide	Formed from
Maltose	Glucose + glucose
Sucrose	Glucose + fructose
Lactose	Glucose + galactose

Each disaccharide is C₁₂H₂₂O₁₁, which is 2 × C₆H₁₂O₆ minus H₂O. Hydrolysis reverses the reaction and gives back the two monosaccharides.

Worked example 2. A 17.1 g sample of sucrose is completely hydrolysed. Use relative molecular masses of 342 for sucrose, 180 for glucose and 18 for water. Find the mass of water used and the total mass of sugars produced.

1. Moles of sucrose = 17.1 ÷ 342 = 0.0500 mol.
2. One sucrose uses one water molecule, so 0.0500 mol of water is used.
3. Mass of water = 0.0500 × 18 = 0.90 g.
4. Mass conserved: sugars produced = 17.1 + 0.90 = 18.0 g, which is 9.0 g of glucose and 9.0 g of fructose.

Polysaccharides

Polysaccharides are formed by the condensation of many glucose units. Glycogen and starch are made from α-glucose. Cellulose is made from β-glucose. They are polymers, but the shape each one takes is what decides its job.

Starch (plant cells)

Starch is the energy store of plants, held as grains in chloroplasts and storage organs. It is a mixture of two polymers of α-glucose.

• Amylose is an unbranched chain that coils into a helix.
• Amylopectin is a chain with side branches.

Why this suits the job: the helix and the branching make a compact molecule, so a lot of glucose is stored in a small space. Starch is insoluble, so it does not change the water potential of the cell or leave it. Branches give many ends where enzymes can hydrolyse off glucose quickly when the plant needs respiratory substrate.

Glycogen (animal cells)

Glycogen is the energy store of animals, found as granules mainly in liver and muscle cells. It is made of α-glucose and is more highly branched than starch, with shorter chains.

Why this suits the job: more branches means more free ends, so glucose can be released by hydrolysis faster than from starch. That matches animals, which are usually more active than plants and need glucose quickly. It is also compact and insoluble.

Worked example 3. A fragment of glycogen contains 40 glucose units with three branch points. Find the number of water molecules released in making it and its relative molecular mass (glucose 180, water 18).

1. A branch point is still just one more glycosidic bond, so bonds = 40 − 1 = 39, and 39 water molecules were released.
2. Mass = (40 × 180) − (39 × 18) = 7200 − 702 = 6498.
3. Complete hydrolysis uses 39 water molecules and gives 40 glucose, a total mass of 7200.

Cellulose (plant cell walls)

Cellulose is a structural carbohydrate. It is made of β-glucose. In the β-form the OH on carbon 1 is above the ring, so each β-glucose has to be rotated 180° relative to its neighbour for the bond to form. The chain is therefore straight and unbranched.

• Many hydrogen bonds form between parallel chains.
• Groups of chains form microfibrils, and these form fibres.

Each hydrogen bond is weak, but there are very many of them, so the fibres have high tensile strength. A plant cell wall made of cellulose resists the pressure of water entering by osmosis, so the cell does not burst. The cell structure revision notes show where the wall fits in the plant cell.

Summary of the three

	Starch	Glycogen	Cellulose
Monomer	α-glucose	α-glucose	β-glucose
Found in	Plant cells	Animal cells	Plant cell walls
Shape	Helix (amylose), branched (amylopectin)	Highly branched	Straight chains, bonded into microfibrils
Function	Energy store	Energy store	Structural strength

Biochemical tests

Benedict’s test for reducing sugars

All monosaccharides and some disaccharides (maltose and lactose) are reducing sugars. Sucrose is non-reducing.

1. Add Benedict’s solution (blue) to the sample.
2. Heat in a hot water bath.
3. Positive result: the blue changes through green, yellow and orange to a brick-red precipitate. More sugar gives a colour further along that sequence. Negative: the solution stays blue.

Test for non-reducing sugars

Start with the ordinary Benedict’s test. A sample that stays blue contains no reducing sugar, so move on:

1. Heat a fresh sample with dilute hydrochloric acid. This hydrolyses the glycosidic bond.
2. Neutralise the acid, for example with sodium hydrogencarbonate, because Benedict’s solution works in alkaline conditions.
3. Add Benedict’s solution and heat. A brick-red precipitate shows a non-reducing sugar was in the sample, because hydrolysis released reducing sugars.

Iodine test for starch

Add iodine in potassium iodide solution. Yellow-brown stays yellow-brown if there is no starch, and turns blue-black if starch is present. No heating is needed.

Worked example 4. Four unknown samples W to Z each contain one of starch, glucose, sucrose or lactose. Identify them.

Sample	Benedict’s alone	Boiled with acid, neutralised, then Benedict’s	Iodine
W	Brick-red	Not needed	Yellow-brown
X	Blue	Brick-red	Yellow-brown
Y	Blue	Blue	Blue-black
Z	Brick-red	Not needed	Yellow-brown

Y is starch (iodine positive, no sugar). X is sucrose (negative at first, positive after hydrolysis). W and Z are both reducing sugars, glucose and lactose. The tests cannot tell these two apart, so another method such as chromatography is needed.

Using the tests quantitatively

The specification suggests two techniques that build on these tests.

Calibration with a dilution series (colorimetry). Make a dilution series from a glucose stock, react each tube identically with Benedict’s solution, and measure the colour of the remaining solution in a colorimeter. Plot a calibration curve. Read off an unknown from it.

Worked example 5. A 20 g dm⁻³ glucose stock is diluted to 10 cm³ of each standard. The absorbance of the solution left after the Benedict’s test falls as sugar rises, because more copper is used up.

Glucose (g dm⁻³)	0	5	10	15	20
Stock (cm³)	0	2.5	5.0	7.5	10.0
Water (cm³)	10.0	7.5	5.0	2.5	0
Absorbance	0.80	0.62	0.45	0.27	0.10

An unknown gives 0.36. It lies between 10 and 15 g dm⁻³. Interpolating: 10 + (0.45 − 0.36) ÷ (0.45 − 0.27) × 5 = 12.5 g dm⁻³.

Chromatography. Spot samples and known standards on a pencil origin line. Run a solvent past them. Sugars move different distances. R_f is the spot distance divided by the solvent-front distance, both measured from the origin line.

Worked example 6. The solvent front moved 8.0 cm. Standards: glucose 3.2 cm, fructose 4.4 cm, galactose 2.0 cm. An unknown shows spots at 3.2 cm and 4.4 cm. R_f values are 0.40, 0.55 and 0.25, so the unknown contains glucose and fructose, as would be expected from hydrolysed sucrose.

Common errors

• Describing starch as a polymer of β-glucose, or cellulose as α-glucose.
• Saying hydrolysis releases water. It uses water.
• Writing n instead of n − 1 for the number of water molecules.
• Calling maltose or lactose non-reducing.
• Forgetting to neutralise before the second Benedict’s test.
• Describing the cellulose bonds between chains as glycosidic. They are hydrogen bonds.
• Saying starch stores energy “because it is insoluble” with no link to osmosis or water potential.

Official syllabus

AQA, AS and A-level Biology specification (7401 and 7402), version 1.5, for exams from June 2016 onwards. Sections 3.1.1 and 3.1.2.