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: C = 12.011, H = 1.008, O = 15.999 g mol⁻¹, so glucose (C₆H₁₂O₆) is 180.156 g mol⁻¹ and water is 18.015 g mol⁻¹. A calculator is assumed. Keep full calculator values until the last step. All data are fictional.
Question 1 (multiple choice · foundation)
Which row correctly describes a polysaccharide?
- (A) Glycogen: found in plants; linear; gives structural support
- (B) Cellulose: found in plant cell walls; linear and unbranched; gives structural support
- (C) Starch: found in animals; highly branched; stores energy
- (D) Cellulose: found in animal liver cells; branched; stores energy
Answer and explanation
Answer: (B). Cellulose chains are straight and unbranched. They lie side by side, hydrogen-bond to each other and form the strong fibres of plant cell walls.
- (A) is wrong because glycogen is a highly branched energy store in animals and fungi.
- (C) swaps the organisms: starch is the plant energy store. The highly branched animal store is glycogen.
- (D) describes glycogen, not cellulose. Animals do not make cellulose.
Question 2 (multiple choice · core)
A plant cell builds an unbranched chain of 250 glucose units by dehydration synthesis. How many water molecules are released?
- (A) 125
- (B) 249
- (C) 250
- (D) 251
Answer and explanation
Answer: (B). One water molecule is released for each link. A chain of 250 units has 250 − 1 = 249 links, so 249 water molecules are released.
- (A) assumes one water per pair of monomers. Each link releases one water, and there are far more links than pairs.
- (C) counts one water per monomer. The first glucose has nothing to join to until the second arrives.
- (D) adds one instead of taking one away.
Question 3 (multiple choice · core)
A glycogen molecule and an unbranched glucose chain contain the same number of glucose units. Which statement best explains why glycogen can release glucose more quickly?
- (A) Branching gives glycogen many more chain ends, and enzymes can remove glucose from many ends at the same time.
- (B) Branching makes glycogen dissolve in the cytoplasm, so glucose units diffuse away from it.
- (C) The links at branch points are hydrogen bonds, which break more easily than covalent bonds.
- (D) Branching packs more glucose units into each molecule, so each enzyme action releases more glucose.
Answer and explanation
Answer: (A). The enzymes that release glucose work at the free ends of chains. More branches mean more ends, so more glucose can be removed per second.
- (B) is wrong: each glucose unit is held by a covalent bond until it is hydrolysed; units do not drift off.
- (C) is wrong because every link between glucose units, including those at branch points, is a covalent bond.
- (D) contradicts the question: both molecules hold the same number of units. Branching changes the shape, not the number of monomers.
Question 4 (diagram · core)
The diagram shows a small, fictional carbohydrate made only of glucose units.
(a) State the number of monomers and the number of covalent links. State whether the molecule is linear or branched. (b) How many water molecules were released when it was built? Give its molecular formula and molar mass. (c) The molecule is fully hydrolysed. State how many water molecules are used, and show that mass is conserved for one mole of the molecule. (d) Using the rule from Worked example 2 of the study guide (one releasing end per chain), how many releasing ends does this molecule have? How many would an unbranched chain of 12 units have?
Worked solution and suggested Marlbridge rubric
(a) 8 + 4 = 12 monomers. 7 links in the main chain, 1 link joining the branch, 3 links within the branch: 11 links. The molecule is branched. Check: any connected chain or branched structure of n units has n − 1 links, and 12 − 1 = 11.
(b) One water per link: 11 water molecules. Formula: 12 × C₆H₁₂O₆ = C₇₂H₁₄₄O₇₂; remove 11 H₂O (22 H, 11 O): C₇₂H₁₂₂O₆₁. Molar mass = 12 × 180.156 − 11 × 18.015 = 2161.872 − 198.165 = 1963.7 g mol⁻¹.
(c) Hydrolysis uses one water per link: 11 water molecules. Mass in: 1963.7 g of carbohydrate + 11 × 18.015 = 198.2 g of water = 2161.9 g. Mass out: 12 × 180.156 = 2161.9 g of glucose. The two match, so mass is conserved.
(d) Two chains (the main chain and one side chain), so 2 releasing ends. An unbranched chain of 12 units is one chain, so 1 releasing end. Same number of units, twice as many places for enzymes to work.
| Point | What earns it |
|---|---|
| 1 | 12 monomers, 11 links and “branched” |
| 1 | 11 waters and C₇₂H₁₂₂O₆₁ |
| 1 | 1963.7 g mol⁻¹ (accept 1964) |
| 1 | 11 waters used in hydrolysis with a correct mass balance |
| 1 | 2 ends compared with 1 for the unbranched chain |
Accept a formula or mass carried forward from an error in the number of links.
Question 5 (constructed response · core)
A student tests whether the enzyme amylase can break down starch and cellulose. She sets up four tubes, each at 37 °C for 20 minutes, with three repeats of each. She then measures the concentration of small sugars (mainly maltose) with a sugar test kit. The kit reads about 0.1 mmol L⁻¹ even when no sugar is present.
| Tube | Contents | Trial 1 | Trial 2 | Trial 3 |
|---|---|---|---|---|
| A | starch + amylase | 4.6 | 5.1 | 4.7 |
| B | starch + boiled amylase | 0.1 | 0.2 | 0.1 |
| C | cellulose + amylase | 0.1 | 0.1 | 0.2 |
| D | cellulose + water | 0.1 | 0.1 | 0.1 |
Small sugars / mmol L⁻¹ after 20 minutes.
(a) Identify the independent variable and the dependent variable. (b) State the purpose of tube B and of tube D. (c) Calculate the mean for tubes A and C. (d) Make a claim about what amylase can do, support it with evidence from the table, and give the reasoning. (e) Another student says: “Amylase had no effect on cellulose because cellulose is not made of glucose.” Evaluate this statement.
Model answer and suggested Marlbridge rubric
(a) Independent variable: the substrate (starch or cellulose), together with whether active amylase is present. Dependent variable: the concentration of small sugars after 20 minutes, in mmol L⁻¹.
(b) Tube B (boiled amylase) tests whether active enzyme is needed: boiling destroys the enzyme’s shape. Tube D is the baseline for cellulose with no enzyme: it shows that cellulose releases no sugar on its own.
(c) A: (4.6 + 5.1 + 4.7) ÷ 3 = 14.4 ÷ 3 = 4.8 mmol L⁻¹. C: (0.1 + 0.1 + 0.2) ÷ 3 = 0.4 ÷ 3 = 0.13, about 0.1 mmol L⁻¹.
(d) Claim: amylase hydrolyses starch but not cellulose. Evidence: tube A averaged 4.8 mmol L⁻¹, about 48 times the kit’s baseline of 0.1, while C stayed at the baseline, the same as D. Reasoning: amylase catalyses hydrolysis of the covalent links between glucose units in starch, releasing small sugars. An enzyme acts only on links that fit its active site, and the links in cellulose are joined in a different way, so amylase cannot break them.
(e) The statement is wrong in its reason. Cellulose is a polymer of glucose, like starch. The difference is in how the glucose units are joined (and cellulose is unbranched), so amylase does not fit those links. The observation (no effect) is correct; the explanation is not.
| Point | What earns it |
|---|---|
| 1 | Independent and dependent variables correct |
| 1 | Purpose of both B and D |
| 1 | Both means correct |
| 1 | Claim supported by comparing A with the baseline and C with D |
| 1 | Reasoning: enzyme fits starch links, not cellulose links; and (e) corrected: cellulose is made of glucose |
You do not need to name the type of linkage. Do not award the last point for “cellulose is too big for the enzyme”: both polymers are large.
Question 6 (constructed response · stretch)
In a fictional strain of mice, a mutation lowers the activity of the enzyme that adds branches to glycogen. Researchers extract liver glycogen from normal and mutant mice and measure, for equal masses of glycogen:
| Measurement | Normal mice | Mutant mice |
|---|---|---|
| Mean chain length / glucose units per chain | 13 | 40 |
| Rate of glucose release by liver enzymes / µmol min⁻¹ mg⁻¹ | 8.4 | 2.9 |
(a) Calculate the percentage decrease in the rate of glucose release in the mutant. (b) For a glycogen molecule of 1300 glucose units, calculate the number of chains for each strain. Compare the ratio of chain numbers with the ratio of release rates. (c) Predict how blood glucose in the mutant mice would change during a period of running, compared with normal mice. Justify the prediction.
Model answer and suggested Marlbridge rubric
(a) (8.4 − 2.9) ÷ 8.4 × 100 = 5.5 ÷ 8.4 × 100 = 65.5% decrease (about 65%).
(b) Normal: 1300 ÷ 13 = 100 chains. Mutant: 1300 ÷ 40 = 32.5, about 33 chains. Ratio of chains: 100 ÷ 32.5 = 3.1. Ratio of rates: 8.4 ÷ 2.9 = 2.9. The two ratios are similar, which supports the idea that the number of chain ends sets the release rate.
(c) During running, muscles take up glucose from the blood, and the liver normally replaces it by breaking down glycogen. In mutant mice, glycogen has fewer branches, so fewer chain ends. Enzymes can remove glucose from fewer places at once, so the liver releases glucose more slowly. Prediction: blood glucose would fall more, or fall sooner, in the mutant mice than in normal mice.
| Point | What earns it |
|---|---|
| 1 | 65.5% (accept 65% or 66%) |
| 1 | 100 and 32.5 (or 33) chains |
| 1 | Both ratios and a comparison linking chain number to rate |
| 1 | Prediction: blood glucose falls more or sooner in mutants |
| 1 | Justification: fewer branches → fewer ends → slower breakdown of glycogen in the liver |
Accept “the mutants tire sooner” if linked to slower glucose release. Do not award the last point for “the mutant has less glucose stored”: the data compare equal masses of glycogen.
Question 7 (constructed response · stretch)
Starch and cellulose are both polymers of glucose.
(a) Describe two differences in structure between starch and cellulose. (b) Explain how the structure of cellulose suits its function in plant cell walls. (c) A student claims: “A cow and a human eating the same mass of grass get the same amount of glucose from its cellulose, because cellulose is just glucose.” Evaluate the claim.
Model answer and suggested Marlbridge rubric
(a) Any two: starch contains branched chains, but cellulose is unbranched (linear); the glucose units are joined in a different way in the two polymers; cellulose chains lie side by side and hydrogen-bond to neighbouring chains, while starch forms compact grains.
(b) Cellulose chains are straight, so many chains can lie close together. Their –OH groups form hydrogen bonds from chain to chain. Each hydrogen bond is weak, but there are very many, so the chains bundle into strong fibres. These fibres resist stretching, so the wall stops the cell from bursting when water enters and helps support the plant.
(c) The claim is not supported. It is true that cellulose is made of glucose. But glucose can only be absorbed after the links between units are hydrolysed. Human digestive enzymes cannot break the links in cellulose, so humans get almost no glucose from it; it passes through as fibre. Cows have microbes in their gut that make enzymes able to break these links, so cows can obtain energy from cellulose, mostly from substances the microbes make as they break the glucose down.
| Point | What earns it |
|---|---|
| 1 | Two correct structural differences |
| 1 | Straight chains lie close together and form hydrogen bonds between chains |
| 1 | Many bonds give strong fibres, linked to support or resisting bursting |
| 1 | Claim rejected: humans lack enzymes to hydrolyse cellulose |
| 1 | Cows rely on gut microbes that make the needed enzymes |
Accept “beta” and “alpha” links if used correctly, but they are not required.
How did you do?
- Q1 or Q7 wrong: re-read “Linear or branched: shape decides the job” in the study guide, and look again at Figure 2.
- Q2 or Q4 counting wrong: rework Worked example 1. Remember: n monomers, n − 1 links, n − 1 waters.
- Q3 or Q6 wrong: rework Worked example 2: more branches, more ends, faster release.
- Q5 incomplete: check that you named a purpose for each control and supported your claim with numbers from the table.
Then tick off the topic checklist.