OCR A Level Biology A: Proteins, inorganic ions and biochemical tests (H420) – 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 follow the OCR A Level Biology A specification (H420), Version 4.1 (April 2026), first assessment 2025, module 2.1.2 Biological molecules, statements (k) to (s). Read the study guide or the revision notes first. Course links: OCR A Level Biology hub and printable checklist.

Questions

1. (a) Name the two functional groups at the ends of an amino acid molecule. [2] (b) Name the element present in the R group of cysteine that is absent from most other amino acids. [1]

2. A polypeptide contains 212 amino acids. (a) State the number of peptide bonds and the number of water molecules released when it formed. [2] (b) Name the type of reaction that forms a peptide bond and describe how the bond is broken. [2]

3. (a) Name the level of protein structure described by the sequence of amino acids. [1] (b) Name the level of structure in which hydrogen bonds between backbone groups form an α-helix. [1] (c) Explain how the tertiary structure of a globular protein is stabilised, naming three bonds or interactions. [3]

4. Haemoglobin is a conjugated protein. (a) Explain what is meant by a conjugated protein, naming the group in haemoglobin. [2] (b) State the level of structure that shows haemoglobin is made of four polypeptides. [1] (c) Explain why haemoglobin is soluble. [2]

5. (a) Name the fibrous protein described in each case: (i) found in hair and nails; (ii) stretches and recoils in artery walls; (iii) gives tendons high tensile strength. [3] (b) Explain why a fibrous protein would be unsuitable for carrying oxygen in blood. [2]

6. Insulin is a globular protein made of two polypeptide chains. (a) State the bond that holds the two chains together. [1] (b) Explain how the shape of insulin relates to its function as a hormone. [3]

7. Give the name and symbol of the ion described. (a) Taken up from soil by plant roots and used to make amino acids. (b) Carries most carbon dioxide in blood plasma. (c) Needed for muscle contraction and bone strength. (d) Found in ATP and in the backbone of nucleic acids. (e) Concentration determines acidity. (f) Released by decay and oxidised by nitrifying bacteria. [6]

8. Five unknown solutions, P to T, were tested. Each shows the colour seen.

Solution	Benedict’s	Benedict’s after boiling with acid and neutralising	Iodine	Biuret	Emulsion
P	orange	orange	yellow-brown	blue	clear
Q	blue	brick-red	yellow-brown	blue	clear
R	blue	orange	blue-black	blue	clear
S	blue	blue	yellow-brown	purple	clear
T	blue	blue	yellow-brown	blue	cloudy white

(a) Identify the main substance in each of P to T. [5] (b) Explain why the solution is neutralised before the second Benedict’s test. [1]

9. Describe how you would use Benedict’s reagent and a colorimeter to compare the reducing sugar content of two fruit juices, C and D. [6]

10. A student measured pigment leaking from beetroot discs. Standards were made from a 100% extract. The absorbances were: 0% 0.00; 20% 0.09; 40% 0.19; 60% 0.28; 80% 0.38; 100% 0.47. (a) Describe how to make 25 cm³ of the 40% standard from the extract. [2] (b) Explain why the colorimeter is zeroed with distilled water. [1] (c) Discs heated to 60 °C gave an absorbance of 0.33. Use the standards to find the extract concentration. [2] (d) Suggest why more pigment leaked from heated discs. [2] (e) Explain why a colorimeter is better than judging colour by eye. [1]

11. A protein was hydrolysed and the amino acids were separated by chromatography on a plate. Spots were 1.6, 3.2, 5.2 and 6.4 cm from the origin. The solvent front was 8.0 cm. Standards had Rf values J 0.20, K 0.40, L 0.65, M 0.80 and N 0.55. (a) Explain why the protein was hydrolysed and name the bond broken. [2] (b) Work out Rf for every spot. [2] (c) Identify the amino acids present and state which standard is absent. [2] (d) Give the reason for using ninhydrin, and the reason for marking the origin in pencil. [2]

Answers

1. (a) Amine group (NH₂) [1] and carboxyl group (COOH) [1]. (b) Sulfur [1]. [3] Examiner insight: Give the group name, not just the element; “amino” alone is less precise than “amine group”.

2. (a) Peptide bonds = 212 − 1 = 211 [1]; water molecules = 211 [1]. (b) Condensation [1]; hydrolysis breaks it by adding water across the peptide bond [1]. [4] Examiner insight: Show the “n − 1” step, since a bare 212 is a common slip.

3. (a) Primary structure [1]. (b) Secondary structure [1]. (c) Hydrogen bonds between polar R groups [1]; ionic bonds between oppositely charged R groups, or disulfide bonds between cysteine residues [1]; hydrophobic interactions pull non-polar R groups into the centre [1]. [5] Examiner insight: Say which parts bond. “Between R groups” matters because the backbone is secondary structure.

4. (a) A globular protein combined with a non-protein prosthetic group [1]; the haem group (containing Fe²⁺) [1]. (b) Quaternary structure [1]. (c) Hydrophilic R groups face outwards [1] and attract water molecules, while hydrophobic groups are buried inside [1]. [5] Examiner insight: Link the outside-facing R groups to water; “it dissolves because it is globular” is too vague.

5. (a) (i) Keratin [1]; (ii) elastin [1]; (iii) collagen [1]. (b) Fibrous proteins are insoluble [1], so they could not dissolve in plasma or be packed in a cell, and they have no binding site for oxygen [1]. [5] Examiner insight: In part (a), a protein name alone is enough; do not add structure details the question does not ask for.

6. (a) Disulfide bond [1]. (b) It is soluble, so it can be carried in blood plasma [1]; its specific three-dimensional shape is complementary to receptors [1] on target cells, so it binds and triggers a response [1]. [4] Examiner insight: “Fits the receptor” is stronger when you name the receptor’s location (target cell membrane).

7. (a) Nitrate, NO₃⁻ [1]. (b) Hydrogencarbonate, HCO₃⁻ [1]. (c) Calcium, Ca²⁺ [1]. (d) Phosphate, PO₄³⁻ [1]. (e) Hydrogen ion, H⁺ [1]. (f) Ammonium, NH₄⁺ [1]. [6] Examiner insight: Write both name and symbol with the correct charge; include the charge, since the symbol is ambiguous without it.

8. (a) P: reducing sugar (positive straight away) [1]. Q: non-reducing sugar, such as sucrose (positive only after hydrolysis) [1]. R: starch (blue-black with iodine; acid hydrolyses it to glucose) [1]. S: protein (purple biuret) [1]. T: lipid (cloudy emulsion) [1]. (b) Benedict’s reagent only works in alkaline conditions, so the acid must be neutralised [1]. [6] Examiner insight: Use the iodine column to separate Q from R, since both are blue then coloured after acid.

9. Add equal volumes of juice C and D to separate tubes [1]; add the same excess volume of Benedict’s reagent and heat both for the same time [1]; filter or centrifuge to remove the precipitate [1]; zero the colorimeter on a blank with a red filter [1]; measure the absorbance of each blue filtrate, repeating and finding a mean [1]; the lower absorbance means more reducing sugar, ideally confirmed with a calibration curve [1]. [6] Examiner insight: Mention control variables explicitly; “same volume, same time, same temperature” is the usual gap.

10. (a) Volume of extract = 0.40 × 25 = 10 cm³ [1]; add 15 cm³ distilled water [1]. (b) It removes the effect of the water and cuvette, so absorbance reflects pigment only [1]. (c) 0.33 lies between 0.28 (60%) and 0.38 (80%): 60 + (0.05 ÷ 0.10) × 20 [1] = 70% [1]. (d) Heat denatures membrane proteins, or disrupts the membrane [1], so more pigment leaks out [1]. (e) It gives an objective measurement of absorbance and detects small differences [1]. [8] Examiner insight: Show a calibration reading with two bracketing values; a bare “70%” is harder to credit if wrong.

11. (a) Hydrolysis breaks peptide bonds [1], releasing amino acids to separate [1]. (b) Rf = spot distance ÷ 8.0: 0.20, 0.40, 0.65, 0.80 [2]. (c) The spots match J, K, L and M [1]; N is absent [1]. (d) Ninhydrin makes colourless amino acids visible, turning them purple [1]; pencil does not dissolve or run in the solvent [1]. [8] Examiner insight: Rf is a ratio, so give no units and keep two decimal places.

Where marks are usually lost

• Giving 212 peptide bonds for 212 amino acids instead of 211.
• Describing quaternary structure as folding of one polypeptide.
• Saying biuret detects amino acids rather than peptide bonds.
• Omitting neutralisation, or the control, in the non-reducing sugar test.
• Writing ion symbols without the charge, or confusing nitrate and nitrite.
• Calling a fibrous protein soluble.
• Forgetting to zero the colorimeter or to say which filter is used.
• Taking spot distances from the plate edge instead of the origin line.
• Not scaling an answer by the dilution factor.

Next steps

• Revision notes
• Study guide
• OCR A Level Biology hub and printable checklist
• All free 10-minute diagnostics
• Book a free trial class

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).