OCR A Level Biology A: Proteins, inorganic ions and biochemical tests (H420)

This guide follows the OCR A Level Biology A specification (H420), Version 4.1 (April 2026), for first assessment in 2025. It covers module 2.1.2 Biological molecules, statements (k) to (s): amino acids, peptide bonds, protein structure, globular and fibrous proteins, inorganic ions, the biochemical tests, colorimetry and chromatography. Water, carbohydrates and lipids (statements (a) to (j)) are taught in separate pages.

Use it with the revision notes and the practice questions. Course links: OCR A Level Biology hub, printable checklist, Foundations in Biology study guide and all free 10-minute diagnostics.

What this unit covers

Statement	Skill to demonstrate
2.1.2(k)	Draw and label the general structure of an amino acid
2.1.2(l)	Explain how dipeptides and polypeptides form and break down by peptide bonds
2.1.2(m)	Describe primary, secondary, tertiary and quaternary structure and the bonds and interactions involved
2.1.2(n)	Relate the structure of globular proteins (haemoglobin, a named enzyme, insulin) to function; explain a conjugated protein
2.1.2(o)	Give properties and functions of collagen, keratin and elastin (no structure details needed)
2.1.2(p)	Recognise the symbols for ten inorganic ions and state their roles
2.1.2(q)	Carry out and interpret biuret, Benedict’s, iodine and emulsion tests
2.1.2(r)	Use quantitative methods, including colorimetry, to find a concentration
2.1.2(s)	Explain and use paper and thin layer chromatography, including Rf values

Amino acids (2.1.2(k))

Proteins contain the elements C, H, O and N, and most also contain S. Each amino acid has a central carbon atom with four different attachments:

• amine group, NH₂
• carboxyl group, COOH
• a single hydrogen
• an R group (a variable side chain)

Written as a diagram: H₂N – CH(R) – COOH. About twenty different amino acids occur in proteins. They differ only in the R group, which can be small, charged, polar, non-polar or sulfur-containing. Cysteine’s R group contains sulfur, which is why it can form disulfide bonds.

Peptide bonds (2.1.2(l))

A condensation reaction joins the carboxyl group of one amino acid to the amine group of the next. One molecule of water is released and a covalent peptide bond (C–N) forms. Joining two amino acids gives a dipeptide; joining many gives a polypeptide. Hydrolysis reverses this: water is added across the peptide bond, which breaks it. Protease enzymes catalyse hydrolysis in digestion.

Worked example 1. Human insulin has an A chain of 21 amino acids and a B chain of 30 amino acids. How many peptide bonds and how many water molecules were released when it formed?

1. A chain: 21 − 1 = 20 peptide bonds.
2. B chain: 30 − 1 = 29 peptide bonds.
3. Total = 49 peptide bonds, so 49 water molecules were released.

One chain of n amino acids contains n − 1 peptide bonds. The order of amino acids can vary hugely: 20 choices at each position gives 20³ = 8000 possible tripeptides and 20⁴ = 160 000 tetrapeptides.

Levels of protein structure (2.1.2(m))

• Primary structure: the sequence of amino acids, held by peptide bonds. A gene determines it.
• Secondary structure: regular folding of the polypeptide backbone into an α-helix or β-pleated sheet, held by hydrogen bonds between the C=O and N–H groups of the backbone.
• Tertiary structure: the overall three-dimensional shape of one polypeptide, formed by interactions between R groups.
• Quaternary structure: two or more polypeptides (and any prosthetic group) held together in one protein.

Interaction	Between which R groups	Notes
Hydrogen bond	Polar R groups	Weak but numerous; broken by heat
Ionic bond	Oppositely charged R groups	Broken by changes in pH
Disulfide bond	Two cysteine residues (S–S)	Covalent, so strong
Hydrophobic interaction	Non-polar R groups	Cluster in the centre, away from water
Hydrophilic interaction	Polar or charged R groups	Face outwards towards water

Shape determines function. If heat or pH change breaks the weak bonds, the tertiary structure unfolds and the protein can no longer do its job.

Globular proteins (2.1.2(n))

Globular proteins are compact and roughly spherical. Hydrophobic R groups point inwards and hydrophilic ones point outwards, so most are soluble. That suits transport, catalysis and signalling.

• A named enzyme, such as amylase. A single polypeptide folded so that a specific active site forms. Its shape is complementary to starch, so it hydrolyses glycosidic bonds.
• Insulin. A hormone made of two polypeptide chains joined by disulfide bonds. It is soluble, so it travels in blood plasma, and its shape fits receptors on target cells.
• Haemoglobin. A conjugated protein: a globular protein combined with a non-protein prosthetic group. It has four polypeptides (quaternary structure), each with a haem group containing an iron(II) ion, Fe²⁺, which binds oxygen. Haemoglobin is soluble, so it can be packed into red blood cells to carry oxygen.

Worked example 2. Haemoglobin has two chains of 141 amino acids and two chains of 146 amino acids. Calculate the number of peptide bonds.

1. Each 141-chain: 140 bonds, so two chains = 280.
2. Each 146-chain: 145 bonds, so two chains = 290.
3. Total = 570 peptide bonds. The haem groups add no peptide bonds.

Fibrous proteins (2.1.2(o))

Fibrous proteins are long, insoluble and have structural roles. You need properties and functions only, not molecular structure.

Protein	Key properties	Functions and locations
Collagen	Very high tensile strength, insoluble, flexible but does not stretch	Tendons, ligaments, skin, bone, cartilage and artery walls
Keratin	Strong, tough, insoluble and rich in sulfur (many disulfide bonds)	Hair, nails, hooves, feathers and the outer layer of skin
Elastin	Stretches and then recoils to its original shape	Walls of arteries, lungs and skin, where tissue must stretch and return

Inorganic ions (2.1.2(p))

Learn the symbol and at least one role for each. Roles beyond these are covered in other modules.

Ion	Role
Calcium, Ca²⁺	Muscle contraction, nerve-ending vesicle release, blood clotting, bone and teeth
Sodium, Na⁺	Nerve impulses; co-transport of glucose and amino acids; water potential
Potassium, K⁺	Nerve impulse recovery; opening of stomata
Hydrogen, H⁺	Determine pH; proton gradients that drive ATP synthesis
Ammonium, NH₄⁺	Form of nitrogen in the nitrogen cycle, converted by bacteria to nitrate
Nitrate, NO₃⁻	Taken up by plants to make amino acids, proteins and nucleotides
Hydrogencarbonate, HCO₃⁻	Carries carbon dioxide in blood; acts as a pH buffer
Chloride, Cl⁻	Balances charge when HCO₃⁻ leaves red blood cells; moves through channels in secretion
Phosphate, PO₄³⁻	Part of ATP, nucleotides and phospholipids; bone
Hydroxide, OH⁻	Makes solutions alkaline; pH rises as OH⁻ increases

Biochemical tests (2.1.2(q))

Test	Reagent and method	Positive result	Negative result
Biuret (protein)	Add sodium hydroxide, then a few drops of dilute copper(II) sulfate	Purple or lilac	Stays blue
Benedict’s (reducing sugar)	Add Benedict’s reagent, heat in a hot water bath	Green, yellow, orange then brick-red precipitate	Stays blue
Iodine (starch)	Add iodine in potassium iodide solution	Blue-black	Stays yellow-brown
Emulsion (lipid)	Shake the sample with ethanol, then add the mixture to water	Cloudy white emulsion	Stays clear

Biuret detects peptide bonds, so it shows proteins and polypeptides but not single amino acids. Benedict’s reagent is reduced by reducing sugars (all monosaccharides, and disaccharides such as maltose and lactose).

Non-reducing sugars (such as sucrose). These give a negative Benedict’s result at first. Use a second portion of the sample:

1. Boil it with dilute hydrochloric acid, which hydrolyses the glycosidic bond.
2. Cool, then neutralise with sodium hydrogencarbonate until fizzing stops.
3. Add Benedict’s reagent and heat.

A positive result now means a non-reducing sugar was present. A negative result in both tests means no sugar. The sequence matters because Benedict’s reagent only works in alkaline conditions.

Quantitative methods and colorimetry (2.1.2(r))

Colour change is only qualitative. A colorimeter measures how much light a coloured solution absorbs, so you can find a concentration.

1. Choose a filter of the complementary colour: a blue solution absorbs red light, so use a red filter.
2. Make standards of known concentration by serial dilution.
3. Zero the colorimeter with a blank (such as distilled water in a matching cuvette).
4. Measure absorbance for each standard and plot a calibration curve.
5. Measure the unknown and read its concentration from the curve.

For Benedict’s reagent, filter off the precipitate and measure the blue filtrate. The more reducing sugar there was, the less blue remains, so absorbance falls as concentration rises.

Worked example 3. Glucose standards (g per 100 cm³) and filtrate absorbances were: 0, 0.92; 0.2, 0.77; 0.4, 0.61; 0.6, 0.47; 0.8, 0.31; 1.0, 0.16. A juice diluted 1 in 10 gave 0.54. Find its glucose concentration.

1. Absorbance 0.54 lies between 0.4 (0.61) and 0.6 (0.47).
2. Fraction of the way: (0.61 − 0.54) ÷ (0.61 − 0.47) = 0.07 ÷ 0.14 = 0.5.
3. Diluted concentration = 0.4 + 0.5 × 0.2 = 0.50 g per 100 cm³.
4. Undiluted: 0.50 × 10 = 5.0 g per 100 cm³.

To make 10 cm³ of the 0.6 standard from a 1.0 stock, mix 6 cm³ stock with 4 cm³ water.

Chromatography (2.1.2(s))

Chromatography separates a mixture. A solvent (the mobile phase) moves through a stationary phase and carries each substance a different distance, which depends on its solubility in the solvent and its attraction to the stationary phase.

• Paper chromatography: the stationary phase is cellulose paper.
• Thin layer chromatography (TLC): the stationary phase is a thin layer of silica gel on a glass or plastic plate. TLC is usually faster and gives sharper separation.

Method: draw the origin line in pencil (ink would dissolve and run), apply small concentrated spots, stand the plate with the origin above the solvent, and remove it when the solvent is near the top. Mark the solvent front at once. Colourless spots need a locating agent such as ninhydrin, which turns amino acids purple, or UV light. Proteins must first be hydrolysed to amino acids. It can analyse proteins, carbohydrates, vitamins and nucleic acids.

Rf = distance moved by the solute ÷ distance moved by the solvent. It has no units and lies between 0 and 1. Compare it with standards run on the same plate.

Worked example 4. Spots lie 2.4, 4.5 and 6.3 cm from the origin; the solvent front is 9.0 cm.

1. Rf values: 2.4 ÷ 9.0 = 0.27; 4.5 ÷ 9.0 = 0.50; 6.3 ÷ 9.0 = 0.70.
2. A standard with Rf 0.27 in an 11.0 cm run would travel 0.27 × 11.0 = 3.0 cm (rearranged: distance = Rf × solvent distance).

Common errors

• Writing that biuret detects amino acids, or that the iodine test gives purple.
• Heating the biuret, iodine or emulsion tests.
• Saying sucrose gives a positive Benedict’s result without hydrolysis.
• Taking spot distances from the plate edge rather than from the origin line.
• Saying the R group forms peptide bonds. It is the amine and carboxyl groups.

Official syllabus

OCR Level 3 Advanced GCE in Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025, published by OCR. Module 2.1.2 Biological molecules, statements (k) to (s).