These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: molar mass of glucose = 180.2 g mol⁻¹; molar mass of water = 18.0 g mol⁻¹. A calculator is assumed. All data sets are fictional.
Question 1 (multiple choice · foundation)
Which statement correctly describes dehydration synthesis between two monomers?
- (A) A water molecule is added; H joins one monomer and –OH joins the other, and the bond between them breaks.
- (B) An H is removed from one monomer and an –OH from the other; a covalent bond forms between the monomers and one water molecule is released.
- (C) Each monomer loses a whole water molecule, and a hydrogen bond forms between them.
- (D) The monomers are dried out, and they stick together by attraction between their partial charges.
Answer and explanation
Answer: (B). One monomer loses H, the other loses –OH. Together these form one water molecule, which is released. The two monomers are joined by a new covalent bond.
- (A) describes hydrolysis, the reverse reaction.
- (C) has two errors: only one water molecule forms per bond, and the link between monomers is covalent, not a hydrogen bond.
- (D) takes “dehydration” to mean drying. The reaction happens in watery cytoplasm, and partial-charge attractions are not covalent bonds.
Question 2 (multiple choice · core)
A straight polymer made of 40 monomers is completely hydrolysed into single monomers. How many water molecules are used?
- (A) 20
- (B) 39
- (C) 40
- (D) 41
Answer and explanation
Answer: (B). A straight chain of n monomers has n − 1 bonds. 40 − 1 = 39 bonds. Each hydrolysis breaks one bond and uses one water molecule, so 39 water molecules are used.
- (A) halves the number of monomers, as if each water molecule released two monomers. Each water molecule breaks only one bond.
- (C) counts one bond per monomer. The first monomer has nothing before it to bond to.
- (D) adds one instead of subtracting one.
Question 3 (multiple choice · core)
Which of the following is an example of dehydration synthesis?
- (A) Enzymes in the small intestine breaking proteins down into amino acids
- (B) A liver cell joining many glucose molecules to form glycogen
- (C) Glucose dissolving in blood plasma as it forms hydrogen bonds with water
- (D) A disaccharide being split into two monosaccharides
Answer and explanation
Answer: (B). Joining monomers (glucose) into a polymer (glycogen) is polymerization by repeated dehydration synthesis, releasing one water per bond.
- (A) is digestion: proteins are broken down by hydrolysis.
- (C) is dissolving. No covalent bonds are made or broken; glucose molecules are only surrounded by water.
- (D) splits a molecule into its monomers, which is hydrolysis.
Question 4 (visual representation · core)
The diagram shows a reaction involving a short polymer.
(a) Name the reaction shown, and give two pieces of evidence from the diagram. (b) Describe what happens to the atoms of each water molecule. (c) Explain why the diagram shows 3 water molecules, not 4. (d) Describe how a diagram of the reverse reaction would differ.
Model answer and suggested Marlbridge rubric
(a) Hydrolysis. Evidence (any two): water is on the reactant side, before the arrow; the bonds between monomers are broken; one large molecule becomes four smaller ones; each product has gained an H or an –OH.
(b) Each water molecule splits at one bond. Its H joins one monomer (for example, H joins monomer 2) and its –OH joins the neighbouring monomer (for example, –OH joins monomer 1). Monomers 2 and 3 sit between two broken bonds, so each gains both an H and an –OH, from two different water molecules.
(c) Four monomers in a chain are joined by 4 − 1 = 3 covalent bonds. One water molecule is used per bond broken, so 3 water molecules are needed.
(d) The arrow would point from the four separate monomers to the chain (dehydration synthesis). The 3 H₂O would be on the product side. The H and –OH groups at the joins would be removed, and three new covalent bonds would form.
| Point | What earns it |
|---|---|
| 1 | Hydrolysis named with two valid pieces of evidence from the diagram |
| 1 | Water split: H to one monomer and –OH to the other |
| 1 | 3 bonds in a 4-unit chain, one water per bond |
| 1 | Reverse: direction reversed and water shown as a product, with 3 bonds formed |
Do not award point 2 for “the water joins the monomers” with no split into H and –OH.
Question 5 (calculation · core)
Use glucose = 180.2 g mol⁻¹ and water = 18.0 g mol⁻¹.
(a) Three glucose molecules join in a straight chain. Calculate the molar mass of the product. (b) A straight chain of glucose units has a molar mass of 1640.0 g mol⁻¹. Calculate the number of glucose units. (c) One mole of the chain in (b) is completely hydrolysed. Calculate the mass of water used, and show that the total mass of glucose produced equals the mass of the chain plus the water.
Worked solution and suggested Marlbridge rubric
(a) n = 3, so 2 bonds and 2 water molecules lost. Mass = 3 × 180.2 − 2 × 18.0 = 540.6 − 36.0 = 504.6 g mol⁻¹.
(b) mass = n × 180.2 − (n − 1) × 18.0 = 162.2n + 18.0. 162.2n = 1640.0 − 18.0 = 1622.0, so n = 1622.0 ÷ 162.2 = 10 glucose units. Check: 10 × 180.2 − 9 × 18.0 = 1802.0 − 162.0 = 1640.0 ✓
(c) 10 units → 9 bonds → 9 mol of water used. Mass of water = 9 × 18.0 = 162.0 g. Mass of glucose produced = 10 × 180.2 = 1802.0 g. Chain + water = 1640.0 + 162.0 = 1802.0 g ✓
| Point | What earns it |
|---|---|
| 1 | 504.6 g mol⁻¹, subtracting 2 (not 3) water molecules |
| 1 | Correct expression for the chain’s mass in terms of n |
| 1 | n = 10, with a check or clear working |
| 1 | 162.0 g of water and the mass balance 1640.0 + 162.0 = 1802.0 shown |
Common errors: 540.6 (forgetting the water) or 486.6 (subtracting 3 water instead of 2) in (a). Accept (c) carried forward from an error in (b).
Question 6 (experiment · core)
Lactase is an enzyme that hydrolyses the disaccharide lactose into glucose and galactose. A student sets up three tubes at 37 °C, each with 10.0 mmol L⁻¹ lactose solution, and measures glucose concentration with a glucose sensor.
- Tube A: lactose + lactase solution
- Tube B: lactose + lactase solution that had been boiled
- Tube C: lactose + the same volume of distilled water
| Time / min | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| Tube A glucose / mmol L⁻¹ | 0.0 | 2.4 | 4.6 | 6.4 | 7.8 |
| Tube B glucose / mmol L⁻¹ | 0.0 | 0.1 | 0.1 | 0.2 | 0.2 |
| Tube C glucose / mmol L⁻¹ | 0.0 | 0.0 | 0.1 | 0.1 | 0.1 |
(a) Identify the independent variable, the dependent variable and one controlled variable. (b) Calculate the mean rate of glucose production in tube A from 0 to 10 min and from 10 to 20 min. Suggest why the rate falls. (c) Calculate the percentage of the lactose hydrolysed in tube A by 20 min, and the concentration of water used up. (d) Explain the purpose of tubes B and C.
Model answer and suggested Marlbridge rubric
(a) Independent variable: whether active enzyme is present (active lactase, boiled lactase, no enzyme). Dependent variable: glucose concentration (mmol L⁻¹), a measure of how much lactose has been hydrolysed. Controlled (any one): temperature (37 °C), starting lactose concentration, total volume, timing of readings.
(b) 0–10 min: (4.6 − 0.0) ÷ 10 = 0.46 mmol L⁻¹ min⁻¹. 10–20 min: (7.8 − 4.6) ÷ 10 = 0.32 mmol L⁻¹ min⁻¹. The rate falls because lactose is being used up: there are fewer lactose molecules left to hydrolyse. (The 5-minute increases, 2.4, 2.2, 1.8 and 1.4, show the same steady slowing.)
(c) Each lactose gives one glucose, so 7.8 mmol L⁻¹ of lactose has been hydrolysed. 7.8 ÷ 10.0 × 100 = 78%. One water is used per bond broken, so 7.8 mmol L⁻¹ of water has been used.
(d) Tube C shows that lactose hardly breaks down in water alone over 20 min, so the glucose in A is not just slow, uncatalysed hydrolysis. Tube B has the same enzyme solution but inactive, so it shows the effect is due to active enzyme, not to something else in the solution (such as glucose already present). Both readings stay at 0.2 mmol L⁻¹ or below.
| Point | What earns it |
|---|---|
| 1 | Independent variable, dependent variable and a valid controlled variable |
| 1 | Both rates correct with units |
| 1 | Falling rate linked to less substrate (lactose) remaining |
| 1 | 78% and 7.8 mmol L⁻¹ of water, linked to one water per bond broken |
| 1 | Roles of both B and C explained |
Do not award point 3 for “the enzyme gets used up”. Enzymes are not consumed in the reaction (Unit 3).
Question 7 (argumentation · stretch)
A student completely hydrolyses 3.424 g of maltose (a disaccharide of two glucose units) in a sealed tube. From the data above, take maltose = 2 × 180.2 − 18.0 = 342.4 g mol⁻¹. She finds that the glucose produced has a mass of 3.604 g. She claims: “The reaction created 0.180 g of new mass.”
(a) Show that 3.604 g is the expected mass of glucose. (b) Evaluate the student’s claim, using what happens in hydrolysis. (c) Predict the mass of product if the same number of glucose molecules were joined back into maltose, and explain.
Model answer and suggested Marlbridge rubric
(a) Moles of maltose = 3.424 ÷ 342.4 = 0.0100 mol. Each maltose gives 2 glucose, so 0.0200 mol of glucose. Mass = 0.0200 × 180.2 = 3.604 g ✓
(b) The claim is wrong. Mass is not created. Hydrolysis uses one water molecule per bond: 0.0100 mol of water = 0.0100 × 18.0 = 0.180 g. That water’s H and –OH become part of the glucose molecules. The extra 0.180 g came from the water already in the tube, so the total mass in the sealed tube does not change.
(c) 3.424 g of maltose. Dehydration synthesis would remove one water per bond formed, so 0.180 g leaves the sugar as water, and 3.604 − 0.180 = 3.424 g.
| Point | What earns it |
|---|---|
| 1 | 0.0100 mol maltose → 0.0200 mol glucose → 3.604 g |
| 1 | Claim rejected: the extra mass is the water used, 0.180 g, with calculation |
| 1 | H and –OH from water added to the products |
| 1 | 3.424 g predicted, linked to water released in dehydration synthesis |
How did you do?
- Q1, Q3 or Q4 wrong: re-read “Dehydration synthesis”, “Hydrolysis” and “Reading diagrams of these reactions” in the study guide, and look again at Figure 1.
- Q2 or Q5 wrong: rework Worked example 1 and draw the chain to count bonds (Figure 2).
- Q6 incomplete: check you linked each glucose to one bond broken and one water used, and that you explained both control tubes.
- Q7 incomplete: re-read Worked example 2 on where the atoms of water go.
Then tick off the topic checklist.