Cambridge International AS & A Level Biology 9700: Proteins – 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.

These questions cover section 2.3 Proteins, part of Topic 2 Biological molecules, in the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027: learning outcomes 1 to 8. It is AS Level content, examined on Papers 1 and 2, and Paper 4 can draw on it. Calculators are allowed in Cambridge 9700 Biology examinations.

Learn it first: Proteins study guide · Recall: Proteins revision notes · Course hub: Cambridge A Level Biology · Checklist: 9700 checklist · More Topic 2 questions: Biological molecules practice

Questions

1. Draw the general structure of an amino acid. Label the amine group, the carboxyl group and the R group. [3]

2. Two amino acids join to form a dipeptide.

(a) Describe how the peptide bond forms. [3] (b) Name the reaction that breaks a peptide bond, and state what is needed for it to happen in a cell. [2]

3. (calculator allowed) The Mr values of three amino acids are glycine 75, alanine 89 and serine 105. The Mr of water is 18.

(a) Calculate the Mr of the tripeptide Gly–Ala–Ser. [2] (b) State how many water molecules are needed to hydrolyse completely a single polypeptide of 250 amino acids. [1]

4. Name the level of protein structure (primary, secondary, tertiary or quaternary) described by each statement.

(a) Two lengths of a polypeptide lie side by side, hydrogen bonded backbone to backbone. [1] (b) The order Met–Lys–Thr–Ala at one end of a chain. [1] (c) A disulfide bond between cysteine 12 and cysteine 95 of the same chain. [1] (d) Three polypeptide chains held together to make one functional protein. [1]

5. Parts of a folded polypeptide bring these R groups close together.

(a) Name the interaction most likely to form between: (i) two phenylalanine R groups (non-polar); (ii) the –OH groups of serine and threonine; (iii) an arginine R group (positively charged) and a glutamate R group (negatively charged); (iv) two cysteine R groups. [4] (b) The protein is placed in a solution of much lower pH. Suggest which of the interactions in (a) is most likely to break, and explain why. [2]

6.

(a) Explain, in terms of R groups, why most globular proteins are soluble in water. [2] (b) Classify haemoglobin, collagen, amylase and keratin as globular or fibrous, and state the type of role each class generally has. [2]

7. Describe the structure of a molecule of haemoglobin. [5]

8. (calculator allowed) In adult haemoglobin each α-globin chain has 141 amino acids and each β-globin chain has 146.

(a) Calculate the number of peptide bonds in one haemoglobin molecule. [2] (b) Taking the Mr of haemoglobin as 64 500 and the Ar of iron as 55.8, calculate the percentage by mass of iron in haemoglobin. Give your answer to 2 significant figures. [2] (c) A red blood cell holds 2.7 × 10⁸ haemoglobin molecules. Calculate the maximum number of oxygen molecules it can carry. [1] (d) Explain why a person whose diet contains too little iron may become breathless during exercise. [2]

9. Describe the structure of a collagen molecule and how collagen molecules are arranged to form collagen fibres. [6]

10. (calculator allowed)

(a) The triple-helical region of each chain in a collagen molecule has 1023 amino acids in a Gly–X–Y repeat. Calculate the number of glycine residues in this region of one collagen molecule. [2] (b) A collagen molecule 300 nm long appears 24 mm long on an electron micrograph. Calculate the magnification. [2] (c) A mutation replaces one of the glycines with an amino acid that has a large R group. Suggest and explain how this could affect the collagen. [3]

11. Compare the structures of haemoglobin and collagen, and explain how each structure suits the protein’s function. [8]

Answers

1. Central carbon bonded to H, –NH₂, –COOH and R, with all bonds shown, including the –OH of the carboxyl group [1]. Amine group (–NH₂) labelled [1]. Carboxyl group (–COOH) and R group both labelled [1]. [3]

Examiner insight: A carboxyl group drawn as –CO₂ or –COO without the hydrogen usually loses the structure mark, even if the labels are right.

2. (a) A condensation reaction [1]. The –OH from the carboxyl group of one amino acid and an –H from the amine group of the other are removed as water [1]. A covalent –CO–NH– bond forms between the carbon of one and the nitrogen of the other [1]. (b) Hydrolysis [1]; water is added, catalysed by a protease enzyme [1].

Examiner insight: In (a), “water is lost” without saying which groups it comes from gains only the first mark; the groups must be named.

3. (a) Two condensations remove 2 × 18 = 36: 75 + 89 + 105 − 36 [1] = 233 [1]. (b) 249 [1].

Examiner insight: Subtracting 18 once instead of twice gives 251 and loses the accuracy mark; the method mark still needs the subtraction to be shown.

4. (a) Secondary [1] (b) Primary [1] (c) Tertiary [1] (d) Quaternary [1]

Examiner insight: (c) catches many candidates: a disulfide bond within one chain is tertiary; it is quaternary only when it links two different chains.

5. (a) (i) Hydrophobic interaction [1]; (ii) hydrogen bond [1]; (iii) ionic bond [1]; (iv) disulfide bond [1]. (b) The ionic bond [1]. Extra H⁺ ions make the negatively charged –COO⁻ of glutamate gain H⁺ and become uncharged –COOH, so there is no longer an attraction between opposite charges [1].

Examiner insight: For (b), naming the ionic bond with no reason scores one mark only; the second mark needs the change in charge on an R group.

6. (a) Hydrophobic (non-polar) R groups are folded inside the molecule [1]; hydrophilic (polar or charged) R groups are on the outside and form interactions with water molecules [1]. (b) Globular: haemoglobin and amylase; fibrous: collagen and keratin [1]. Globular proteins generally have physiological roles and fibrous proteins structural roles [1].

Examiner insight: In (a), “globular proteins are polar” is not enough; the mark needs the positions of hydrophobic and hydrophilic R groups.

7.

• Globular protein with four polypeptide chains, so it has quaternary structure [1]
• Two α-globin and two β-globin chains [1]
• One haem group per chain, four in total, each held in a pocket of the folded chain [1]
• Each haem contains an iron ion, Fe²⁺ [1]
• Chains held together by hydrophobic interactions, hydrogen bonds and ionic bonds; hydrophilic R groups on the outside [1]

Examiner insight: “Four haem groups” and “one haem group per chain” are the same point and score once; you need five different ideas for five marks.

8. (a) Total amino acids = 2 × 141 + 2 × 146 = 574; four separate chains, so 574 − 4 [1] = 570 [1]. (b) 4 × 55.8 ÷ 64 500 × 100 [1] = 0.35% [1]. (c) 4 × 2.7 × 10⁸ = 1.08 × 10⁹ [1]. (d) Iron is needed to make haem, so less haemoglobin is made and less oxygen is carried in the blood [1]; muscles receive less oxygen for aerobic respiration, so breathing rate rises [1].

Examiner insight: In (a), subtracting 1 instead of 4 gives 573: treating the molecule as one chain loses the accuracy mark. In (b), using one iron atom instead of four is the most common slip.

9.

• Three polypeptide chains, each a helix, wound into a triple helix [1]
• Hydrogen bonds between the chains [1]
• Glycine at almost every third position, small enough to let the chains pack tightly [1]
• Molecules lie side by side, with staggered ends [1]
• Covalent cross-links between R groups (lysine) of neighbouring molecules [1]
• Molecules form fibrils, which bundle into fibres [1]

Examiner insight: Calling the chains “α-helices” is a contradiction that can cost a mark; describe them as helices wound into a triple helix.

10. (a) 1023 ÷ 3 = 341 glycines per chain [1]; × 3 chains = 1023 [1]. (b) Convert to the same units: 24 mm = 24 000 000 nm [1]; 24 000 000 ÷ 300 = × 80 000 [1]. (c) The large R group cannot fit in the centre of the triple helix [1], so the three chains cannot pack closely and the triple helix is disrupted [1]; the collagen fibres have lower tensile strength and tissues such as tendon or bone are weaker [1].

Examiner insight: In (b), magnification has no units; writing “80 000 nm” loses the accuracy mark. In (c), each mark needs a link: shape, then packing, then strength.

11.

• Haemoglobin is globular, with four chains (2 α, 2 β) folded into a compact shape; collagen is fibrous, with three chains wound into a long triple helix [1]
• Haemoglobin has haem with Fe²⁺, which binds O₂ reversibly; collagen has no prosthetic group [1]
• Each haemoglobin carries up to four O₂, for oxygen transport [1]
• Hydrophilic R groups outside make haemoglobin soluble, so it is carried at high concentration in red blood cells [1]
• Binding of one O₂ changes the shape of haemoglobin, so others bind more easily [1]
• Collagen: glycine every third amino acid gives tight packing of the chains [1]
• Collagen molecules are staggered and covalently cross-linked into fibrils and fibres [1]
• So collagen has high tensile strength and is insoluble, suited to tendons, skin and bone [1]

Examiner insight: A comparison needs matching points (“four chains but three chains”); two separate descriptions with no link can lose comparison marks, and every structural feature should be paired with its function.

Where marks are usually lost

• Drawing amino acids with missing hydrogens on the amine group or the –OH missing from the carboxyl group.
• Saying water comes from the R groups during condensation.
• Forgetting to subtract one water per peptide bond in Mr calculations, or subtracting one too few.
• Counting peptide bonds for a multi-chain protein as if it were one chain.
• Classing a disulfide bond inside one chain as quaternary structure.
• Naming “bonds” without saying which: hydrogen, ionic, disulfide or hydrophobic.
• Using one iron, or one O₂, per haemoglobin molecule.
• Giving magnification with units, or failing to convert mm to nm first.
• Describing collagen strength without cross-links or staggered ends.

Next steps

• Proteins revision notes
• Proteins study guide
• Transport in mammals study guide for haemoglobin and oxygen transport
• Cambridge A Level Biology hub
• 9700 printable checklist
• Free 9700 AS diagnostic
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment – Topic 2, Biological molecules (section 2.3 Proteins).