OxfordAQA A-Level Chemistry: Amino acids and proteins (9620)

This study guide teaches section 3.3.13, Amino acids and proteins, from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020. It covers all three subsections: 3.3.13.1 Amino acids, 3.3.13.2 Proteins and 3.3.13.3 Action of anticancer drugs. The section belongs to the International A2 organic content: all of it is International A-level only. Unit 4 (Organic 2 and Physical 2) assesses sections 3.3.7 to 3.3.16; Unit 5 (Practical and synoptic) can draw on any content.

Pair it with the revision notes and the practice questions. Hydrogen bonding comes from shapes, polarity and intermolecular forces, and the square planar shape of cisplatin from transition metals. For the whole course, see the OxfordAQA A-Level Chemistry hub, the printable checklist and the free diagnostics.

What this section covers

Section	What you must know or do	Stage
3.3.13.1	Amino acids have acidic and basic properties and form zwitterions; draw the zwitterion and the ions in acid and in alkaline solution	International A-level only
3.3.13.2	Proteins are amino acids joined by peptide links; draw peptides of up to three amino acids and the amino acids formed by hydrolysis	International A-level only
3.3.13.2	Primary, secondary (α-helix and β-pleated sheet) and tertiary structure; identify each in diagrams; explain how hydrogen bonds and S–S bonds maintain these structures	International A-level only
3.3.13.2	Separate and identify amino acids by thin-layer chromatography; locate them with ninhydrin or ultraviolet light; calculate Rf values	International A-level only
3.3.13.3	Cisplatin, a Pt(II) complex, prevents DNA replication by ligand replacement; explain adverse effects and the balance of benefit and harm	International A-level only

Section 3.3.13 lists no required practical activity. Thin-layer chromatography returns in section 3.3.16. In 9620 this section has no outcomes on enzymes or on DNA structure; DNA enters only through cisplatin.

3.3.13.1 Amino acids

Each amino acid here has an amino group, –NH₂, and a carboxyl group, –COOH, on the same carbon: H₂N–CH(R)–COOH. The side chain R makes each amino acid different.

The –COOH group is acidic: it can lose H⁺ to become –COO⁻. The –NH₂ group is basic: the lone pair on nitrogen accepts H⁺ to give –NH₃⁺. Inside one molecule, the carboxyl group can pass H⁺ to the amino group. The result is a zwitterion: one species that carries a positive charge and a negative charge, with an overall charge of zero.

H₂N–CH(R)–COOH   →   ⁺H₃N–CH(R)–COO⁻

What happens in acid and in alkali

• Acid solution. There is plenty of H⁺. The –COO⁻ end takes H⁺ back and becomes –COOH. The –NH₃⁺ end stays. Overall charge +1.
• Alkaline solution. There is plenty of OH⁻. The –NH₃⁺ end loses H⁺ to OH⁻ and becomes –NH₂. The –COO⁻ end stays. Overall charge −1.

Check the side chain every time. A second –COOH or –NH₂ in R reacts in the same way as the main one.

Worked example 1: phenylalanine

Phenylalanine is H₂N–CH(CH₂C₆H₅)–COOH. Draw it as a zwitterion, as the ion in acid solution and as the ion in alkaline solution.

1. The side chain, –CH₂C₆H₅, has no acidic or basic group.
2. Zwitterion: move H⁺ from carboxyl to amino. ⁺H₃N–CH(CH₂C₆H₅)–COO⁻, overall charge 0.
3. Acid: protonate the carboxylate. ⁺H₃N–CH(CH₂C₆H₅)–COOH, overall charge +1.
4. Alkali: deprotonate the –NH₃⁺. H₂N–CH(CH₂C₆H₅)–COO⁻, overall charge −1.

Worked example 2: a side chain that reacts

Glutamic acid is HOOC–CH₂CH₂–CH(NH₂)–COOH. Draw the ions formed in an excess of acid and in an excess of alkali.

1. Count the groups: one –NH₂ and two –COOH.
2. Excess acid: the –NH₂ becomes –NH₃⁺; both –COOH stay. HOOC–CH₂CH₂–CH(NH₃⁺)–COOH, charge +1.
3. Excess alkali: both –COOH become –COO⁻; the –NH₂ stays. ⁻OOC–CH₂CH₂–CH(NH₂)–COO⁻, charge −2.

3.3.13.2 Proteins

Peptide links

A carboxyl group on one amino acid condenses with the amino group on a second. Water is lost and a peptide link forms:

     O    H
     ‖    |
 ––C––N––        written in a line as –CO–NH–

This is a condensation reaction: two amino acids give a dipeptide and one H₂O; three give a tripeptide and two H₂O. A protein is a long chain built this way. Write a peptide with the free –NH₂ on the left, so Thr–Leu–Met has threonine at the –NH₂ end.

Worked example 3: drawing a tripeptide

Draw Thr–Leu–Met and calculate its Mr. Threonine is H₂N–CH(CH(OH)CH₃)–COOH (Mr 119.0). Leucine is H₂N–CH(CH₂CH(CH₃)₂)–COOH (Mr 131.0). Methionine is H₂N–CH(CH₂CH₂SCH₃)–COOH (Mr 149.1).

1. Write the three in order: Thr, then Leu, then Met.
2. Link threonine’s C=O to leucine’s N, then leucine’s C=O to methionine’s N.

H₂N–CH(CH(OH)CH₃)–CO–NH–CH(CH₂CH(CH₃)₂)–CO–NH–CH(CH₂CH₂SCH₃)–COOH

3. Check: one free –NH₂, one free –COOH, two –CO–NH– links, side chains unchanged.
4. Two waters lost: Mr = 119.0 + 131.0 + 149.1 − (2 × 18.0) = 363.1 (C₁₅H₂₉N₃O₅S).

Met–Leu–Thr has the same Mr but differs.

Hydrolysis of the peptide link

Breaking the peptide link with water regenerates the amino acids it came from. Water adds across each C–N bond: the carbon gets –OH and the nitrogen gets –H. The peptide is usually heated under reflux with hydrochloric acid or aqueous sodium hydroxide, and the conditions decide which form you draw.

• In excess acid, each amino group is protonated: ⁺H₃N–CH(R)–COOH.
• In excess alkali, each carboxyl group is deprotonated: H₂N–CH(R)–COO⁻.

Worked example 4: products of hydrolysis

The dipeptide H₂N–CH(CH₂C₆H₅)–CO–NH–CH(CH(CH₃)CH₂CH₃)–COOH (Mr 278.0) is heated under reflux with excess hydrochloric acid. Draw the organic products.

1. Find the one peptide link and break the C–N bond.
2. Left fragment: C=O gains –OH, giving phenylalanine. Right fragment: N gains –H, giving isoleucine.
3. Excess acid protonates both amino groups:
   • ⁺H₃N–CH(CH₂C₆H₅)–COOH
   • ⁺H₃N–CH(CH(CH₃)CH₂CH₃)–COOH

Protein structure

Level	Meaning	Held by
Primary	The order of amino acids along the chain	Covalent peptide links
Secondary	Regular local folding of the backbone into an α-helix or a β-pleated sheet	Hydrogen bonds between peptide links
Tertiary	The overall three-dimensional shape of the whole chain	Interactions between side chains, including hydrogen bonds and S–S bonds

Hydrogen bonding. In each peptide link, the N–H hydrogen is δ+ (nitrogen is more electronegative) and the C=O oxygen is δ− with lone pairs. A lone pair on the O of one peptide link attracts the δ+ H of another.

• α-helix: the chain coils, and a C=O on one turn hydrogen bonds to an N–H a few residues further along.
• β-pleated sheet: stretches of chain lie side by side, and C=O groups on one stretch hydrogen bond to N–H groups on the next.

Sulfur–sulfur bonds. The side chain of cysteine is –CH₂SH. Two such –SH groups that meet in the folded chain can be oxidised to an S–S link: a covalent bond, much stronger than a hydrogen bond, that ties together distant parts of the chain and locks the tertiary shape. Methionine’s –CH₂CH₂SCH₃ cannot do this, as it has no S–H bond.

Worked example 5: reading diagrams

Name the level of structure each diagram shows.

Diagram shows	Level	Reason
Three-letter codes in a row, such as Ser–Cys–Leu–Lys	Primary	It is only the sequence
A spiral ribbon with dashed lines between C=O on one turn and N–H on the next	Secondary (α-helix)	Regular folding held by backbone hydrogen bonds
Two arrows lying side by side, joined by dashed lines	Secondary (β-pleated sheet)	Neighbouring stretches linked by backbone hydrogen bonds
A whole chain folded into a compact blob, with an S–S bridge between two distant points	Tertiary	Overall 3D shape fixed by side-chain links

Separating amino acids by TLC

1. Draw a pencil baseline and put small spots of the hydrolysate and of reference amino acids on it.
2. Place the plate in a covered tank with the solvent level below the baseline.
3. Lift the plate out while the solvent is still short of the top edge, and mark the front at once.
4. The spots are colourless. Spray with ninhydrin and warm: amino acids show as purple spots. Or shine ultraviolet light on a plate containing a fluorescent substance: the spots appear dark.
5. Calculate Rf values and compare with the standards.

Rf = (baseline-to-spot distance) ÷ (baseline-to-solvent-front distance).

Rf depends on the balance between solubility in the moving solvent and retention by the solid, so it changes with solvent and plate. Compare only with standards run under the same conditions, ideally on the same plate.

Worked example 6: identifying the spots

A hydrolysis mixture is run beside standards. The solvent front is 5.6 cm above the baseline. The mixture gives spots at 1.4 cm, 3.08 cm and 4.2 cm. Under the same conditions, threonine has Rf 0.25, methionine 0.55 and leucine 0.75. Identify the amino acids.

1. Spot 1: 1.4 ÷ 5.6 = 0.25, which matches threonine.
2. Spot 2: 3.08 ÷ 5.6 = 0.55, which matches methionine.
3. Spot 3: 4.2 ÷ 5.6 = 0.75, which matches leucine.

Rf is a ratio with no units, always between 0 and 1. TLC shows which amino acids are present, not their order.

3.3.13.3 Action of anticancer drugs

Cisplatin, cis-[Pt(NH₃)₂Cl₂], is a platinum(II) complex used to treat some cancers. Its shape is square planar, with the two chloride ligands next to each other.

How it stops DNA replication

The specification describes a ligand replacement reaction with DNA: a chloride ligand on platinum is replaced, and a nitrogen atom uses its lone pair to form a co-ordinate bond to platinum. The specification’s wording places that nitrogen “on a protein”; in the DNA itself, it is usually on a guanine base. Because the two chloride ligands are cis, both can be replaced, so one platinum can bind to two sites close together. The DNA is distorted, and it cannot be copied normally. Without replication, the cancer cell cannot divide.

Adverse effects and the balance of risk

Cisplatin does not pick out cancer cells. It attaches to the DNA of healthy cells as well, which halts their division too, and its side effects include kidney damage and hearing loss. Society has to weigh the benefit, such as shrinking a tumour, against the harm.

Common errors

• Giving glutamic acid in alkali a charge of −1. Both carboxyl groups lose H⁺, so it is −2.
• Linking C to O, or N to N, when drawing a peptide. The link is always C(=O)–N(H).
• Subtracting one water for a tripeptide. Two links means two waters.
• Saying secondary structure is held by S–S bonds. S–S bonds hold the tertiary structure.
• Measuring spot distances from the bottom edge of the plate instead of the baseline.
• Saying cisplatin bonds to DNA through chlorine. The chloride ligands leave.

Next, try the practice questions, or check recall with the revision notes.

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards (first teaching 2019): section 3.3.13 Amino acids and proteins, subsections 3.3.13.1 to 3.3.13.3, International A2 content.