Before you start
This review joins the seven topics of Unit 1. Do the unit diagnostic first and close its gaps. Then try the questions without looking at the answers.
Big ideas of the unit
- Polarity is the starting point. Water’s polar O–H bonds let it form hydrogen bonds (Topic 1.1). Later topics reuse the idea: polar and charged groups mix with water; nonpolar groups are pushed together.
- Elements point to molecules. All four classes contain C, H and O. Sulfur points to proteins, phosphorus to phospholipids and nucleic acids, and nitrogen to nucleic acids and proteins (Topic 1.2).
- One pair of reactions builds and breaks everything. Dehydration synthesis releases one water per link; hydrolysis uses one water per link (Topic 1.3). The rule covers sugars, fats, peptide bonds and the DNA backbone.
- Same monomer, different job. Starch, glycogen and cellulose are all glucose. Branching and linkage decide whether the polymer is a fast store, a slower store or a strong fibre (Topic 1.4).
- Lipids are defined by how they behave with water. Nonpolar C–H bonds make them hydrophobic. Fats store energy densely; amphipathic phospholipids form a bilayer by themselves (Topic 1.5).
- Sequence carries information. DNA’s backbone repeats, so information lies in the base order. Antiparallel strands pair through hydrogen bonds, and nucleotides join only at the 3′ end (Topic 1.6).
- Sequence decides shape; shape decides function. Backbone hydrogen bonds make helices and sheets; R-group interactions fold the chain; several chains may fit together (Topic 1.7).
- Strong bonds hold the sequence, weak attractions hold the shape. Heat can separate DNA strands or unfold a protein without breaking the covalent chain.
Key relationships and methods
| Idea | Rule or method | Where it appears |
|---|---|---|
| Links in a chain | n building blocks in a straight or branched chain → n − 1 links → n − 1 waters released or used | polysaccharides, polypeptides, one DNA strand |
| Fat | glycerol + 3 fatty acids → 3 links, 3 waters | triglycerides |
| Mass of a polymer | n × (monomer mass) − (n − 1) × 18.0 | any chain of identical monomers |
| Energy store | mass = energy ÷ energy per gram (fat 37, carbohydrate 17 kJ g⁻¹) | comparing fat and starch or glycogen |
| Saturation | each C=C removes 2 H and adds a kink; more kinks → more liquid | fatty acids, membrane fluidity |
| Base pairing | A–T (2 hydrogen bonds), C–G (3); strands antiparallel; read 5′ → 3′ | DNA; A–U in RNA |
| Elements | C, H, O in all; S → protein; P → phospholipid or nucleic acid; N → nucleic acid or protein | identifying unknowns, tracers |
| Structure → function | change → which interaction is lost or gained → which level → shape → function | proteins, membranes |
These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not official scoring guidelines. All data sets, molecules and organisms are fictional.
Question 1 (multiple choice · mixed)
When a newly made enzyme folds in the cytoplasm, most of its nonpolar R groups end up clustered in the centre. Which property of water is most directly responsible?
- (A) Water’s high specific heat keeps the protein cool enough to fold.
- (B) Water molecules hydrogen-bond with each other and with polar and charged groups, so nonpolar R groups are left together, away from water.
- (C) Water forms covalent bonds with polar R groups and pulls them outward.
- (D) Surface tension squeezes the chain into a ball.
Answer and explanation
Answer: (B). Nonpolar R groups have no partial charges for water to attract. Water stays hydrogen-bonded to itself and to the polar and charged groups, so the nonpolar groups cluster in the core (hydrophobic interactions).
- (A) names a real property of water, but it does not decide which R groups go inside.
- (C) confuses attraction with covalent bonding. Water attracts polar groups; it does not bond covalently to them.
- (D) misuses surface tension, which is a property of a water surface, not of water around a molecule.
Question 2 (multiple choice · mixed)
A double-stranded DNA fragment of 20 base pairs contains 12 C–G pairs and 8 A–T pairs. It is broken down completely into single nucleotides. How many water molecules are used?
- (A) 19
- (B) 38
- (C) 40
- (D) 90
Answer and explanation
Answer: (B). Each strand of 20 nucleotides has 20 − 1 = 19 covalent links, and each link needs one water to hydrolyse. Two strands: 2 × 19 = 38. The hydrogen bonds between the strands also break, but separating bases does not use water.
- (A) counts one strand only.
- (C) counts one water per nucleotide (2 × 20).
- (D) adds the 52 hydrogen bonds (12 × 3 + 8 × 2) to 38, as if breaking a hydrogen bond were a hydrolysis.
Question 3 (multiple choice · mixed)
Seed X stores 74 kJ of energy as oil. Seed Y stores the same energy as starch. Use 37 kJ g⁻¹ for fat and 17 kJ g⁻¹ for carbohydrate. Which statement is correct?
- (A) The oil store is lighter, 2.0 g against about 4.4 g, because oil is mostly C–H bonds with very little oxygen.
- (B) The oil store is lighter, 2.0 g against about 4.4 g, because oil is a polymer with more links per gram.
- (C) The starch store is lighter, about 2.0 g against 4.4 g, because starch is branched and compact.
- (D) Both stores have the same mass, because they hold the same energy.
Answer and explanation
Answer: (A). Oil: 74 ÷ 37 = 2.0 g. Starch: 74 ÷ 17 = 4.35 g, about 4.4 g, so starch needs about 2.2 times the mass. Fat tails are packed with C–H bonds and contain little oxygen, so each gram releases more energy.
- (B) has the right numbers but a wrong reason: fats are not polymers.
- (C) swaps the two stores. Branching affects how fast glucose is released, not how much energy each gram holds.
- (D) ignores the different energy per gram.
Question 4 (constructed response · mixed)
A student is given four purified samples. Her results:
| Sample | Elements detected | In water | Products of complete hydrolysis |
|---|---|---|---|
| P | C, H, O | dissolves easily | cannot be hydrolysed further; identical to the only product of S |
| Q | C, H, O | does not dissolve; forms oily droplets | glycerol and fatty acids in the ratio 1 : 3 |
| R | C, H, O, N, P | forms sheets two molecules thick | glycerol, two fatty acids and a phosphate-containing group |
| S | C, H, O | forms insoluble granules | glucose only; one molecule gives 2000 glucose |
(a) Identify each sample as a class of molecule. Give one piece of evidence for each. (b) Explain, in terms of polarity, why P dissolves in water but Q does not. (c) Explain why R forms sheets two molecules thick. (d) S is branched. Calculate how many water molecules are used to hydrolyse one molecule of S completely, and explain why branching does not change the answer.
Worked solution and suggested Marlbridge rubric
(a) P: a monosaccharide (glucose): only C, H and O, soluble, cannot be split further, and the same molecule as the monomer of S. Q: a fat (triglyceride): glycerol to fatty acids 1 : 3, and hydrophobic. R: a phospholipid: contains phosphorus, and gives glycerol, two fatty acids and a phosphate group. S: a storage polysaccharide (glycogen or starch): a polymer that gives only glucose.
(b) P has many polar –OH groups that form hydrogen bonds with water, so it dissolves. Q is mostly nonpolar C–H bonds. Water molecules stay hydrogen-bonded to each other instead, so Q is pushed into droplets.
(c) R is amphipathic: a hydrophilic phosphate head and two hydrophobic tails. The heads face the water on both sides and the tails point towards each other, hidden from water. This bilayer forms without any energy input.
(d) 2000 − 1 = 1999 water molecules. Every glucose after the first joins with exactly one new link. A branch changes where a unit joins, not how many links there are.
| Point | What earns it |
|---|---|
| 1 | P and S identified with evidence |
| 1 | Q and R identified, with the 1 : 3 ratio for Q and phosphorus or two fatty acids for R |
| 1 | P dissolves through hydrogen bonds with polar –OH groups; Q is nonpolar so water excludes it |
| 1 | Bilayer explained by hydrophilic heads facing water and hydrophobic tails hidden |
| 1 | 1999 waters, with the reason that branching does not change the number of links |
Total: 5 points. Accept “starch” or “glycogen” for S.
Question 5 (constructed response · mixed)
Portin is a fictional protein that sits in the plasma membrane. One stretch of 20 amino acids crosses the bilayer. Seventeen of its R groups are nonpolar. Researchers made two variants, each with one change at position 9 of this stretch (normally nonpolar). They measured the percentage of Portin molecules found correctly inserted in the membrane.
| Protein | Change at position 9 | Inserted / %: trial 1, 2, 3 |
|---|---|---|
| Wild type | none | 94, 92, 96 |
| Variant A | nonpolar → ionic | 31, 35, 27 |
| Variant B | nonpolar → a different nonpolar | 93, 90, 95 |
(a) Calculate the mean for each protein. (b) Explain why the membrane-crossing stretch has mostly nonpolar R groups. (c) Explain the result for variant A. Name the level of protein structure first changed. (d) State the purpose of variant B, and what its result shows. (e) A student claims: “One amino acid out of 20 cannot matter.” Evaluate the claim using the data.
Worked solution and suggested Marlbridge rubric
(a) Wild type: (94 + 92 + 96) ÷ 3 = 94.0%. Variant A: (31 + 35 + 27) ÷ 3 = 31.0%. Variant B: (93 + 90 + 95) ÷ 3 = 92.7%.
(b) The middle of the bilayer is the phospholipids’ hydrophobic fatty acid tails. Nonpolar R groups can sit against these tails; polar or charged R groups are stable only next to water.
(c) The primary structure changes first: one amino acid is replaced. The ionic R group carries a full charge, so it is attracted to water and cannot sit among the nonpolar tails. The stretch is far less likely to stay in the bilayer. Insertion falls by (94.0 − 31.0) ÷ 94.0 × 100 = 67%.
(d) Variant B controls for the act of changing position 9. Its mean (92.7%) is close to the wild type, and the ranges overlap (90–95 against 92–96). So the type of R group matters, not the change itself.
(e) The claim is not supported. Variant A changes 1 of 20 amino acids, yet insertion falls from 94.0% to 31.0%, with no overlap in the ranges. Variant B also changes 1 amino acid, with almost no effect. Location and R-group properties matter, not the number changed.
| Point | What earns it |
|---|---|
| 1 | All three means correct |
| 1 | Nonpolar R groups linked to the hydrophobic tails in the middle of the bilayer |
| 1 | Variant A: charged R group attracted to water, so the stretch leaves or fails to enter the bilayer; primary structure changed first |
| 1 | Variant B as a control, comparing the ranges or means |
| 1 | Claim rejected using both variants as evidence |
Total: 5 points.
Question 6 (constructed response · mixed)
During digestion, an animal hydrolyses three molecules from its food: an unbranched chain of 400 glucose units, one triglyceride, and a single polypeptide of 150 amino acids. Use glucose = 180.2 g mol⁻¹ and water = 18.0 g mol⁻¹.
(a) Calculate the number of water molecules used for each molecule and in total. (b) Calculate the molar mass of the glucose chain, and show that hydrolysing one mole of it gives a mass of glucose equal to the chain plus the water used. (c) Name the products of each hydrolysis. State which products contain nitrogen, and where. (d) The animal later builds its own glycogen and proteins from these products. Describe the reaction it uses, and what happens to the water count.
Worked solution and suggested Marlbridge rubric
(a) Glucose chain: 400 − 1 = 399. Triglyceride: 3, one per fatty acid. Polypeptide: 150 − 1 = 149. Total 399 + 3 + 149 = 551 water molecules.
(b) Chain: 400 × 180.2 − 399 × 18.0 = 72 080 − 7182 = 64 898 g mol⁻¹. Water used: 399 × 18.0 = 7182 g. Chain + water = 64 898 + 7182 = 72 080 g = 400 × 180.2 g of glucose ✓. The H and –OH of each water join the products.
(c) Chain → glucose. Triglyceride → glycerol and three fatty acids. Polypeptide → amino acids. Only the amino acids contain nitrogen, in the amine group (–NH₂).
(d) Dehydration synthesis: for each link, an –OH leaves one building block and an H the other, as one water molecule. A chain of n units releases n − 1 waters, the same number hydrolysis used.
| Point | What earns it |
|---|---|
| 1 | 399, 3 and 149 |
| 1 | Total 551 |
| 1 | 64 898 g mol⁻¹ with the mass balance shown |
| 1 | Products named and nitrogen placed in the amine group of amino acids only |
| 1 | Dehydration synthesis described, with one water released per link |
Total: 5 points. Accept (b) carried forward from an error in (a).
Question 7 (constructed response · mixed)
A student heats three samples to 95 °C for 5 minutes, then cools them slowly to 25 °C. Sample 1 is a solution of a double-stranded DNA fragment. Sample 2 is a solution of a fictional enzyme. Sample 3 is a variant of the same enzyme with two extra disulfide bridges.
| Sample | Before heating | At 95 °C | After cooling |
|---|---|---|---|
| DNA: % double-stranded | 100 | 0 | 96 |
| Enzyme: activity / units | 60.0 | 0 | 4.2 |
| Variant enzyme: activity / units | 58.8 | 0 | 21.0 |
(a) Identify the bonds that break when the DNA is heated, and explain why each strand stays intact. (b) Explain why most of the DNA becomes double-stranded again on cooling. (c) Calculate the percentage of activity each enzyme keeps after cooling. (d) Explain why the enzyme loses activity at 95 °C. State which level of structure is not changed. (e) Use the data to support a claim about the effect of the extra disulfide bridges.
Worked solution and suggested Marlbridge rubric
(a) The hydrogen bonds between complementary bases on opposite strands break. Each strand is held together by covalent links between the phosphate of one nucleotide and the sugar of the next, which do not break at 95 °C.
(b) Each strand keeps its base sequence. On slow cooling, A meets T and C meets G again, and hydrogen bonds re-form between the complementary strands.
(c) Enzyme: 4.2 ÷ 60.0 × 100 = 7.0%. Variant: 21.0 ÷ 58.8 × 100 = 35.7%.
(d) Heat breaks the weak attractions that hold the shape: backbone hydrogen bonds (secondary structure) and hydrogen bonds, hydrophobic and ionic interactions between R groups (tertiary structure). The protein denatures, the active site loses its shape, and activity falls to zero. The primary structure is unchanged: peptide bonds stay intact. Most chains do not refold correctly on cooling.
(e) Claim: the extra disulfide bridges help the enzyme regain its working shape after heating. Evidence: the variant keeps 35.7% against 7.0%, about five times as much, with almost the same starting activity (58.8 against 60.0 units). Reasoning: disulfide bridges are covalent, so they survive 95 °C and hold distant parts of the chain together, helping it fold back towards its tertiary structure. With one trial each, the result needs repeating.
| Point | What earns it |
|---|---|
| 1 | Hydrogen bonds between bases break; covalent backbone bonds do not |
| 1 | Re-pairing explained by complementary sequences and hydrogen bonds re-forming |
| 1 | 7.0% and 35.7% |
| 1 | Loss of activity linked to weak interactions breaking and changed shape, with primary structure unchanged |
| 1 | Claim about disulfide bridges with both numbers compared |
| 1 | Reasoning: bridges are covalent and survive heating, so they help keep or restore the shape |
Total: 6 points.
How did you do?
Add up your points: 3 for the multiple-choice questions and 21 for the constructed responses. Use the total only to compare with your own next attempt; it does not predict an exam score.
- Lost points on counting or mass (Questions 2, 4(d), 6): go back to the n − 1 rule in Topic 1.3.
- Lost points on polarity and water (Questions 1, 4(b)–(c), 5(b)): reread Topic 1.1 and Topic 1.5.
- Lost points on predictions (Questions 5, 7): practise the chain “change → interaction → level → shape → function” from Topic 1.7.
- Want a quick recheck? Retake the unit diagnostic.
Then tick off each topic checklist: