A Level Biology: Biological Molecules — Practice Questions (Cambridge 9700)

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available free from your board.

Related: Biological molecules study guide and revision notes

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Section A

1. State the positive result and the substance tested for in the Biuret test and the emulsion test. [4]

2. Explain why a control is necessary when carrying out a biochemical test on an unknown sample. [2]

3. Name the four bonds or interactions that can stabilise a protein’s tertiary structure. [4]

Section B

4. A student tests an unknown solution: it gives a brick-red precipitate with Benedict’s test after heating, but no colour change with iodine solution.

(a) State what each result indicates about the solution. [2] (b) Explain why testing with both reagents gives more information than either test alone. [2]

5. Compare the structure of a triglyceride and a phospholipid, and explain why only one of them can form the bilayer structure of a cell membrane. [4]

6. Starch does not give a positive result with Benedict’s test unless it has first been boiled with dilute hydrochloric acid. Explain why. [3]

7. Explain, in terms of hydrogen bonding, why water has a high specific heat capacity, and state one biological consequence of this property. [3]

8. A polypeptide chain folds into a specific tertiary structure. Explain how R-group interactions between amino acids at different points along the chain contribute to this folding. [3]

9. Cellulose and glycogen are both polysaccharides built from glucose monomers, but have very different biological roles. Explain how their structures suit them to structural support and energy storage respectively. [4]

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Answers

1. Biuret test: purple/lilac colour [1], tests for proteins [1]. Emulsion test: white emulsion forms [1], tests for lipids [1].

2. A control confirms that any colour change or precipitate observed is caused specifically by the substance being tested for, rather than by the reagent itself reacting under the test conditions [2].

3. Ionic bonds, hydrogen bonds, disulfide bonds, and hydrophobic interactions between R groups [4] (one mark each).

4. (a) The brick-red precipitate with Benedict’s test indicates the presence of a reducing sugar [1]. The absence of a colour change with iodine indicates the absence of starch [1]. (b) Testing with both reagents together allows a more specific conclusion than either test alone — here, the combined results indicate a reducing sugar is present but starch is not, rather than leaving open the possibility that either or both could be present, as a single test result would [2].

5. Both molecules share a glycerol backbone [1]. A triglyceride has three fatty acid tails, making it wholly non-polar and hydrophobic — an efficient, compact energy store that does not interact with water [1]. A phospholipid replaces one fatty acid tail with a phosphate group, giving it a polar, hydrophilic head alongside its remaining non-polar, hydrophobic tails [1]. Only the phospholipid can form a membrane bilayer, because in water its hydrophilic heads orient outward and hydrophobic tails orient inward, spontaneously forming a stable bilayer — a triglyceride has no hydrophilic region to orient outward in this way [1].

6. Starch is a polysaccharide, and Benedict’s test only detects reducing sugars, not polysaccharides directly [1]. Boiling starch with dilute hydrochloric acid hydrolyses the glycosidic bonds joining the glucose units together [1], breaking the starch down into free glucose (a reducing sugar), which can then give a positive result with Benedict’s test [1].

7. Water molecules are held together by extensive hydrogen bonding between neighbouring molecules [1]. A relatively large amount of energy must be absorbed to break these hydrogen bonds and raise the temperature of water, which is why water has a high specific heat capacity [1]. Biological consequence: this property helps buffer aquatic habitats (or organisms) against rapid temperature change [1].

8. R groups at different points along the polypeptide chain interact with each other as the chain folds — for example, oppositely charged R groups can form ionic bonds, polar R groups can form hydrogen bonds, and non-polar R groups can cluster together through hydrophobic interactions away from surrounding water [2]. These interactions pull specific parts of the chain into a stable three-dimensional arrangement, and because the pattern of R groups is set by the primary sequence, the resulting tertiary structure is determined directly by that sequence [1].

9. Cellulose is built from beta-glucose monomers joined into long, straight, unbranched chains, which run parallel to each other and are cross-linked by many hydrogen bonds to form strong microfibrils suited to providing structural support in plant cell walls [2]. Glycogen is built from alpha-glucose monomers joined into a highly branched structure, which provides many end points for enzymes to rapidly add or remove glucose units, making it well suited to being a readily accessible energy store that can be quickly mobilised or replenished [2].

A note on exam technique for this topic

Every question above rewards linking a structural detail to its functional or observable consequence, rather than stating a fact in isolation — this mirrors the structure-to-function habit the revision notes for this topic identify as the single highest-value skill across Biological molecules. When practising further questions from past papers, check your own answers specifically for this pairing: a correct fact about bonding or structure that is never connected to what it explains functionally is unlikely to gain full marks, even where the underlying knowledge is accurate.