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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

The questions below test section 3.3.13, Amino acids and proteins (3.3.13.1 to 3.3.13.3), of 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 is International A2 content, so International A-level only. Unit 4 (Organic 2 and Physical 2) assesses it, and Unit 5 (Practical and synoptic) can draw on any content.

Work through the study guide and the revision notes before you start. See also the course hub, the printable checklist and the free diagnostics. Question 7 builds on transition metals.

Data. Mr values: threonine 119.0, leucine 131.0, isoleucine 131.0, glutamic acid 147.0, methionine 149.1, phenylalanine 165.0, water 18.0.

Amino acid	Structure
Threonine	H₂N–CH(CH(OH)CH₃)–COOH
Leucine	H₂N–CH(CH₂CH(CH₃)₂)–COOH
Isoleucine	H₂N–CH(CH(CH₃)CH₂CH₃)–COOH
Glutamic acid	H₂N–CH(CH₂CH₂COOH)–COOH
Methionine	H₂N–CH(CH₂CH₂SCH₃)–COOH
Phenylalanine	H₂N–CH(CH₂C₆H₅)–COOH
Lysine	H₂N–CH(CH₂CH₂CH₂CH₂NH₂)–COOH

Questions

1. Three species formed from methionine are shown.

• Species A: ⁺H₃N–CH(CH₂CH₂SCH₃)–COO⁻
• Species B: H₂N–CH(CH₂CH₂SCH₃)–COO⁻
• Species C: ⁺H₃N–CH(CH₂CH₂SCH₃)–COOH

For each species, state whether it is the zwitterion, the ion present in acid solution or the ion present in alkaline solution, and give its overall charge. [3]

2. Draw the structure of the ion formed when lysine is dissolved in an excess of aqueous sodium hydroxide, and state its overall charge. [2]

3. A tripeptide has the sequence isoleucine–methionine–glutamic acid, with isoleucine carrying the free –NH₂ group.

(a) Draw the structure of the tripeptide. [2] (b) Calculate its relative molecular mass. [1] (c) Explain why the tripeptide glutamic acid–methionine–isoleucine is a different compound, even though it has the same Mr. [1]

4. The peptide shown is fully hydrolysed by refluxing it with excess NaOH(aq).

H₂N–CH(CH(OH)CH₃)–CO–NH–CH(CH₂C₆H₅)–CO–NH–CH(CH₂CH(CH₃)₂)–COOH

(a) State the number of peptide links in the molecule. [1] (b) Name the amino acid at the end with the free amino group. [1] (c) Draw the three organic ions present after hydrolysis. [3]

5. After a dipeptide has been hydrolysed, its products are run on a thin-layer chromatography plate. The solvent front ends 6.5 cm above the baseline. Two spots are found at 1.30 cm and 4.55 cm above the baseline. Under the same conditions, standards gave these Rf values: threonine 0.20, methionine 0.45, phenylalanine 0.60, leucine 0.70.

(a) Explain why a developing agent is needed, and describe how ninhydrin is used to locate the spots. [2] (b) Work out Rf for both spots. [2] (c) Name the two amino acids present. [1] (d) Explain why the standards should be run on the same plate as the mixture. [1] (e) Another sample gives a faint extra spot at Rf 0.33 that matches no standard. Suggest one explanation. [1]

6. Identify the level of protein structure shown in each diagram described.

(a) A chain folded into a compact shape, with a bridge between two distant cysteine units. [1] (b) The sequence Met–Thr–Leu–Phe–Glu written out in a line. [1] (c) Two stretches of chain side by side, with dashed lines joining C=O on one to N–H on the other. [1]

7. Cisplatin is cis-[Pt(NH₃)₂Cl₂].

(a) State the shape of cisplatin and the oxidation state of platinum in it. [1] (b) Explain why cisplatin stops cancer cells from replicating their DNA. [3] (c) Explain why drugs such as cisplatin can have adverse effects. [2] (d) Suggest why a doctor may still decide to use cisplatin despite these adverse effects. [1]

8. This question is about hydrogen bonds and S–S bonds in proteins.

(a) Explain how hydrogen bonds hold a β-pleated sheet together. [3] (b) A small protein contains three methionine units and no cysteine. Explain whether S–S bonds can help to maintain its tertiary structure. [2]

9. A dipeptide has the structure H₂N–CH(CH₂CH(CH₃)₂)–CO–NH–CH(CH₂CH₂COOH)–COOH.

(a) Calculate the relative molecular mass of the dipeptide. [1] (b) The dipeptide is dissolved in an excess of cold dilute sodium hydroxide, which does not hydrolyse the peptide link. Draw the ion formed and give its charge. [2] (c) Another sample is heated under reflux with an excess of concentrated hydrochloric acid. Draw the two organic species formed. [2]

10. An unknown tripeptide T has Mr 397.1. A sample of T is fully hydrolysed, and the amino acids are separated by TLC. The solvent front moves 7.0 cm from the baseline, and spots appear at 1.40 cm, 3.15 cm and 4.20 cm from the baseline. Use the Rf standards from question 5, which were obtained under the same conditions.

(a) Calculate the three Rf values and identify the amino acids. [3] (b) Show that your answer to (a) is consistent with the Mr of T. [1] (c) Partial hydrolysis of T gives two dipeptides. In the one with Mr 266.0, threonine carries the free –NH₂ group; in the one with Mr 296.1, phenylalanine does. Deduce the sequence of T, explaining your reasoning. [3] (d) Draw the structure of T. [1]

Answers

1. A: zwitterion, charge 0 [1]. B: ion in alkaline solution, charge −1 [1]. C: ion in acid solution, charge +1 [1]. [3] Examiner insight: Give both the condition and the charge for each species; a correct label with the wrong sign is unlikely to gain credit.

2. H₂N–CH(CH₂CH₂CH₂CH₂NH₂)–COO⁻ [1]. Both –NH₂ groups stay uncharged and only the carboxyl group loses H⁺, so the overall charge is −1 [1]. [2] Examiner insight: An –NH₃⁺ on the side chain cannot survive excess OH⁻, so a structure showing one is unlikely to gain credit.

3. (a) H₂N–CH(CH(CH₃)CH₂CH₃)–CO–NH–CH(CH₂CH₂SCH₃)–CO–NH–CH(CH₂CH₂COOH)–COOH: correct side chains in the right order [1]; two –CO–NH– links with free –NH₂ and –COOH at the ends [1]. (b) 131.0 + 149.1 + 147.0 − (2 × 18.0) = 391.1 [1]. (c) The amino acids are joined in a different order, so the free –NH₂ is now on glutamic acid and the free –COOH on isoleucine [1]. [4] Examiner insight: Glutamic acid’s side-chain –COOH must remain a –COOH in the peptide; using it to make the link changes the structure.

4. (a) 2 [1]. (b) Threonine [1]. (c) H₂N–CH(CH(OH)CH₃)–COO⁻ [1]; H₂N–CH(CH₂C₆H₅)–COO⁻ [1]; H₂N–CH(CH₂CH(CH₃)₂)–COO⁻ [1]. [5] Examiner insight: With excess alkali the carboxyl groups must be drawn as –COO⁻; neutral amino acids answer a different question.

5. (a) Amino acids are colourless, so the spots are invisible [1]. Spray with ninhydrin and warm: purple spots appear [1]. (b) 1.30 ÷ 6.5 = 0.20 [1]; 4.55 ÷ 6.5 = 0.70 [1]. (c) Threonine and leucine [1]. (d) Rf depends on conditions such as the solvent and plate, so standards on the same plate meet exactly the same conditions [1]. (e) Hydrolysis was incomplete, so some dipeptide remains (or an amino acid not among the standards is present) [1]. [7] Examiner insight: Two separate points are needed in (a): why locating is needed and how ninhydrin works; describing only the purple colour leaves one point unmade.

6. (a) Tertiary [1]. (b) Primary [1]. (c) Secondary (β-pleated sheet) [1]. [3] Examiner insight: In (c), hydrogen bonds between backbone C=O and N–H groups mean secondary structure; links between side chains would point to tertiary.

7. (a) Square planar, with platinum in the +2 oxidation state [1]. (b) Cisplatin undergoes a ligand replacement reaction in which a chloride ligand is replaced [1]. A nitrogen atom in the DNA (on a guanine base) uses its lone pair to form a co-ordinate bond to platinum [1]. The DNA is bent out of shape, so replication is blocked and the cancer cell stops dividing [1]. (c) It is not selective and also binds to DNA in healthy cells [1], stopping them dividing and causing side effects such as kidney damage or hearing loss [1]. (d) The benefit of treating the cancer is judged to outweigh the harm from the side effects [1]. [7] Examiner insight: In (b), “platinum bonds to DNA” is too vague for credit; name the reaction type, the nitrogen atom and the effect on replication.

8. (a) Hydrogen bonds link neighbouring stretches of chain lying side by side [1]. Each δ+ hydrogen of an N–H group on one stretch [1] bonds to a lone pair on a δ− carbonyl oxygen in the neighbouring stretch [1]. (b) No: S–S bonds form between –SH groups, as in cysteine side chains [1]. Methionine’s –CH₂CH₂SCH₃ has no S–H bond, so it cannot form one [1]. [5] Examiner insight: “Contains sulfur” is not enough reasoning; the answer must turn on the presence or absence of an –SH group.

9. (a) 131.0 + 147.0 − 18.0 = 260.0 [1]. (b) H₂N–CH(CH₂CH(CH₃)₂)–CO–NH–CH(CH₂CH₂COO⁻)–COO⁻, with both carboxyl groups deprotonated [1]; overall charge −2 [1]. (c) ⁺H₃N–CH(CH₂CH(CH₃)₂)–COOH [1] and ⁺H₃N–CH(CH₂CH₂COOH)–COOH [1]. [5] Examiner insight: In (b), forgetting the side-chain –COOH gives a charge of −1; check the side chain before you total the charges.

10. (a) 1.40 ÷ 7.0 = 0.20; 3.15 ÷ 7.0 = 0.45; 4.20 ÷ 7.0 = 0.60 [1]. Matching with the standards: 0.20 is threonine, 0.45 is methionine [1], and 0.60 is phenylalanine [1]. (b) 119.0 + 149.1 + 165.0 − (2 × 18.0) = 397.1, which matches the Mr of T [1]. (c) 119.0 + 165.0 − 18.0 = 266.0 (threonine + phenylalanine); 165.0 + 149.1 − 18.0 = 296.1 (phenylalanine + methionine) [1]. With the free –NH₂ groups given, these are Thr–Phe and Phe–Met, so phenylalanine sits between the other two [1]. The sequence is Thr–Phe–Met [1]. (d) H₂N–CH(CH(OH)CH₃)–CO–NH–CH(CH₂C₆H₅)–CO–NH–CH(CH₂CH₂SCH₃)–COOH [1]. [8] Examiner insight: Part (b) asks you to show consistency, so set out every Mr and both water subtractions before comparing with 397.1.

Where marks are usually lost

• Ignoring a side-chain –COOH (glutamic acid) or –NH₂ (lysine) when working out an ion and its charge.
• Drawing hydrolysis products in the neutral form when the question states acid or alkali.
• Explaining secondary structure with “hydrogen bonds” but not saying which atoms take part.
• Treating any sulfur-containing side chain as able to form S–S bonds.
• Measuring Rf distances from the bottom edge, or inverting the fraction.
• Explaining cisplatin without ligand replacement or the nitrogen atom.

Next steps

• Amino acids and proteins revision notes
• Amino acids and proteins study guide
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

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.