Chemistry • Year 12 • Module 7 • Lesson 15b
Carbohydrate Biomolecules
Build Band 5–6 extended-response technique on condensation–hydrolysis reasoning, starch–cellulose comparison, and cross-biomolecule analysis in investigative and HSC-style contexts.
1. Data + scenario: complete hydrolysis of a starch fragment (Band 5–6)
8 marks Band 5–6
Scenario. A biochemist isolates a starch fragment containing exactly 10 glucose units linked by α-1,4-glycosidic bonds. She completely hydrolyses it using excess dilute HCl at 100 °C. A second starch fragment of 10 glucose units is hydrolysed by amylase at 37 °C, which first produces maltose molecules and then fully converts them to glucose.
| Measurement | Starch fragment (10 glucose units) |
|---|---|
| Molecular formula of each glucose monomer | C⁶H₁₂O₆ |
| Number of glycosidic bonds in the fragment | 9 |
| Water molecules consumed in complete hydrolysis | ? |
| Number of glucose molecules produced | ? |
| Molecular formula of the starch fragment (before hydrolysis) | ? |
Q1. Analyse and evaluate the scenario above. In your response you must:
- Complete the three missing cells in the table, showing your working for the molecular formula calculation.
- Write the overall balanced equation for the complete hydrolysis of the starch fragment, using the formula you calculated.
- Explain why amylase produces maltose as an intermediate rather than immediately releasing individual glucose molecules, with reference to bond specificity.
- Compare the two hydrolysis methods (acid vs amylase) in terms of specificity, conditions, and the range of products that could be produced.
- State whether complete hydrolysis of cellulose would yield the same result as complete hydrolysis of starch of the same number of glucose units, and explain why or why not.
2. Experimental design — identifying an unknown carbohydrate from hydrolysis products (Band 5–6)
7 marks Band 5–6
Research question. A student is given an unlabelled white powder that is a pure carbohydrate. They believe it may be sucrose, maltose, or starch. Design a scientific investigation to identify which carbohydrate it is by using hydrolysis and chromatographic analysis of the products.
Constraints: Standard lab equipment is available: dilute H₂SO₄, water bath at 100 °C, chromatography paper, appropriate solvent, reference standards of glucose, fructose, and galactose, and iodine solution. The investigation must be completed in a single laboratory session (3 hours).
Q2. Design the investigation in the format below.
- State a hypothesis (include the expected products for each possible identity and how you would distinguish between them).
- Describe the procedure in at least five numbered steps, including the hydrolysis method and how you will use chromatography to identify the monosaccharide products.
- Explain the role of the iodine solution as an additional test for starch (before hydrolysis).
- Create a results table showing what you would expect to observe for each possible identity of the unknown powder.
- State one limitation of using acid hydrolysis in this investigation and suggest an improvement using enzymes instead.
Q1 — Sample Band 6 response (8 marks), annotated
Completed table:
- Water molecules consumed = 9 (one per glycosidic bond; n−1 = 10−1 = 9) [1].
- Glucose molecules produced = 10 [1].
- Molecular formula of starch fragment: 10 × C⁶H₁₂O₆ − 9 × H₂O = C⁶₀H₁ᾔO₅₁ − H₁₈O₉ = C⁶₀H₁ᾌO₅ᾕ. Working: 10 C⁶ = C⁶₀; 10 H₁₂ − 9 × 2H = 120H − 18H = 102H; 10 O₆ − 9O = 60O − 9O = 51O. Formula: C⁶₀H₁₀₂O₅₁ [1 mark for correct formula with working shown; accept minor arithmetic errors if method is correct].
Balanced hydrolysis equation: C⁶₀H₁₀₂O₅₁ + 9H₂O → 10C⁶H₁₂O₆ [1 mark for correctly balanced equation matching the formula calculated].
Amylase and maltose intermediate: Amylase cleaves α-1,4-glycosidic bonds but does so in a sequential manner, releasing the disaccharide maltose as an intermediate product (two glucose units still bonded together). To release individual glucose molecules from maltose, a separate enzyme (maltase) is required. This is because each enzyme is specific to the bond at a particular position along the chain; amylase cleaves internal α-1,4 bonds but not the terminal bond in the same step as producing free glucose [1].
Acid vs amylase comparison: Acid hydrolysis is non-specific — it cleaves any glycosidic bond (both α and β configurations) given sufficient time and temperature; it produces only glucose from any glucose polymer. Amylase is enzyme-specific: it cleaves only α-1,4-glycosidic bonds, does not cleave β-1,4 bonds (in cellulose), and operates under mild conditions (37 °C, neutral pH) in the body. Acid requires elevated temperature (100 °C) and acidic conditions [1 mark for comparing specificity; 1 mark for comparing conditions].
Starch vs cellulose complete hydrolysis: Yes, both would yield the same final product — glucose (C⁶H₁₂O₆) — if completely hydrolysed, because both are glucose polymers. However, cellulose requires cellulase (or concentrated strong acid under harsh conditions) to cleave β-1,4-glycosidic bonds, whereas starch requires amylase (or dilute acid) to cleave α-1,4 bonds. The same number of glucose units and the same number of water molecules would be consumed, giving the same formula and the same products [1].
Marking criteria (8 marks): 1 = water molecules (9); 1 = glucose produced (10); 1 = correct formula with working; 1 = balanced hydrolysis equation; 1 = amylase–maltose intermediate with bond specificity reasoning; 1 = acid vs amylase specificity; 1 = acid vs amylase conditions; 1 = starch vs cellulose comparison (same products, different enzyme / conditions needed).
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis: If the unknown powder is sucrose, complete acid hydrolysis will yield glucose and fructose in equal amounts (detectable as two spots on chromatography). If it is maltose, only glucose will be detected. If it is starch, only glucose will be detected but the powder will give a blue-black colour with iodine solution before hydrolysis (starch is a polysaccharide; sucrose and maltose will not give a positive iodine test) [1].
Procedure: (1) Dissolve 1.0 g of the unknown powder in 20 mL of distilled water in a conical flask. (2) Add 3 drops of iodine solution to 1 mL of the solution; record whether a blue-black colour forms (positive = starch) or no colour change (negative = not starch). (3) Add 5 mL of 1.0 mol/L H₂SO₄ to the remaining solution, heat in a water bath at 100 °C for 30 minutes to completely hydrolyse any glycosidic bonds. Neutralise with NaHCO₃ solution. (4) Apply 2 µL of the hydrolysate and reference standards (glucose, fructose, galactose) side by side onto chromatography paper using a microcapillary. Develop in a suitable solvent system (e.g. butanol–acetic acid–water). (5) Allow the chromatogram to dry, spray with ninhydrin or use a developing reagent suitable for sugars; identify spots by Rf value compared with standards. Record the number and identity of monosaccharide spots [1 per clear step, max 2 marks; bonus mark if iodine test is described as step (2) correctly].
Iodine test role: Iodine (I₂ in KI) interacts with the helical structure of starch, inserting into the coiled amylose chain to produce an intense blue-black colour. Sucrose and maltose (disaccharides) do not have a helical polymer structure, so they give no colour change. A positive iodine test before hydrolysis confirms the sample is starch (a polysaccharide) [1].
Expected results table:
| Unknown identity | Iodine test | Chromatography spots |
|---|---|---|
| Sucrose | Negative (no colour) | 2 spots: glucose + fructose |
| Maltose | Negative (no colour) | 1 spot: glucose only |
| Starch | Positive (blue-black) | 1 spot: glucose only |
[1 mark for a correctly structured results table distinguishing all three possibilities].
Limitation and enzymatic improvement: Acid hydrolysis is non-specific — it could degrade the monosaccharides further under harsh conditions, producing artefacts, and cannot distinguish starch from maltose by products alone. Improvement: use specific enzymes — invertase (cleaves sucrose only), maltase (cleaves maltose only), or amylase (cleaves α-1,4 bonds in starch) — in separate test tubes. Enzyme specificity means each enzyme will only hydrolyse its substrate, giving a positive reaction only for the matching carbohydrate [1 mark for limitation; 1 mark for enzymatic improvement].
Marking criteria (7 marks): 1 = hypothesis identifying expected products for all three possibilities including iodine test prediction; 1 = at least five clear, sequenced procedural steps; 1 = iodine test role explained (starch only gives blue-black due to helical structure); 1 = correct results table distinguishing all three carbohydrates; 1 = valid limitation of acid hydrolysis; 1 = enzyme-based improvement with reasoning; 1 = precise chemical terminology throughout (chromatography, Rf, glycosidic bond, invertase/amylase/maltase, polysaccharide).