Biological Molecules — A-Level Biology Study Guide

For: A-Level Biology candidates sitting A-Level Biology.

Covers: Polar water properties, structure and function of carbohydrates, lipids, proteins, and core enzyme models as required for the A-Level Biology syllabus.

You should already know: IGCSE Biology, basic chemistry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Biology style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Biological Molecules?

Biological molecules are carbon-containing organic compounds that form the structural and functional building blocks of all living organisms, carrying out roles from energy storage to metabolic catalysis. This topic accounts for 8-12% of total marks across A-Level Biology papers 1 (multiple choice), 2 (AS structured), and 4 (A2 structured), and is a prerequisite for all subsequent cell biology, physiology, and biochemistry content in the syllabus. It is also frequently tested in practical assessments (Papers 3 and 5) via qualitative food tests. Common synonyms for this topic include biomolecules and core macromolecules.

2. Water — properties from polarity

Water (${\text{H}}_2\text{O}$) is a polar covalent molecule: oxygen has higher electronegativity than hydrogen, so it pulls shared electrons closer to its nucleus, creating a partial negative charge (${\delta }^{-}$) on the oxygen atom and partial positive charges (${\delta }^{+}$) on both hydrogen atoms. Weak electrostatic hydrogen bonds form between the ${\delta }^{+}$ hydrogen of one water molecule and the ${\delta }^{-}$ oxygen of an adjacent molecule. While individual hydrogen bonds are weak, large numbers of them give water unique biologically critical properties:

1. Cohesion and adhesion: Cohesion is the attraction between adjacent water molecules, while adhesion is the attraction of water to charged surfaces. Together, these allow continuous columns of water to be pulled up plant xylem from roots to leaves without breaking, even against gravity.
2. High specific heat capacity (SHC): Water has an SHC of $4200{\text{ J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}$, meaning it requires large amounts of heat to raise its temperature. This buffers temperature changes in cells and aquatic habitats, keeping internal organism temperatures and ocean environments stable.
3. High latent heat of vaporisation: Water requires $2260{\text{ kJ kg}}^{-1}$ of energy to evaporate, so sweating and transpiration remove excess heat from organisms when water evaporates, acting as a natural cooling mechanism.
4. Universal solvent: Polar and ionic compounds dissolve easily in water, as water molecules surround charged solute particles to form hydration shells. This makes water the primary transport medium in blood, xylem, and phloem, and the medium for all metabolic reactions in the cytoplasm.

Worked Example

If 2 kg of water absorbs $33600\text{ J}$ of heat energy, what is its temperature change? Use the formula $Q=mc\Delta T$, rearranged to $\Delta T=\frac{Q}{mc}$: $$\Delta T=\frac{33600}{2\times 4200}=4^{\circ }\text{C}$$ This small temperature change for a large energy input demonstrates how water’s high SHC stabilises cell temperatures during heat-releasing metabolic reactions.

3. Carbohydrates — mono, di, polysaccharides

Carbohydrates have the general formula ${\text{C}}_x({\text{H}}_2\text{O}{)}_y$, and are grouped by their polymer length:

1. Monosaccharides: Single sugar units, all reducing sugars. The most common is glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$), which has two isomers: $\alpha$-glucose (hydroxyl group on carbon 1 sits below the ring structure) and $\beta$-glucose (hydroxyl group on carbon 1 sits above the ring structure). Examiners regularly test correct drawing of these isomers, so always label carbon 1 in your diagrams.
2. Disaccharides: Two monosaccharides joined by a glycosidic bond, formed via a condensation reaction that releases one water molecule. Hydrolysis reactions break glycosidic bonds by adding a water molecule. Common examples include maltose (two $\alpha$-glucose, 1,4 glycosidic bond), sucrose ($\alpha$-glucose + fructose, non-reducing), and lactose (glucose + galactose).
3. Polysaccharides: Long polymers of monosaccharide subunits, adapted for storage or structural roles:

• Starch: Plant energy storage, a mix of amylose (straight unbranched chains of $\alpha$-glucose with 1,4 glycosidic bonds, helical compact shape) and amylopectin (branched chains with 1,4 and 1,6 glycosidic bonds, multiple free ends for rapid hydrolysis to release glucose when needed).
• Glycogen: Animal energy storage, more highly branched than amylopectin, allowing even faster glucose release for high-energy demand in muscle and liver cells.
• Cellulose: Plant cell wall structural component, made of $\beta$-glucose subunits joined by 1,4 glycosidic bonds. Each $\beta$-glucose is flipped relative to the next, forming straight unbranched chains cross-linked by hydrogen bonds to form strong microfibrils that give plant cells tensile strength.

4. Lipids — triglycerides and phospholipids

Lipids are non-polar organic compounds insoluble in water, soluble in organic solvents such as ethanol. The two core types tested in A-Level Biology are:

1. Triglycerides: Structure consists of one glycerol molecule covalently bonded to three fatty acid molecules via ester bonds, formed by condensation reactions that release 3 water molecules total. Fatty acids are classified as saturated (no carbon-carbon double bonds, straight chains, solid at room temperature, e.g., animal lard) or unsaturated (one or more carbon-carbon double bonds, kinked chains, liquid at room temperature, e.g., olive oil). Key functions include long-term energy storage (39 kJ of energy per gram, twice the energy density of carbohydrates), thermal insulation under the skin, and buoyancy for aquatic organisms such as whales.
2. Phospholipids: Modified triglycerides where one fatty acid is replaced by a charged phosphate group attached to a polar head group (e.g., choline). This makes phospholipids amphipathic: the phosphate head is hydrophilic (attracted to water) and the two fatty acid tails are hydrophobic (repelled by water). In aqueous environments, phospholipids spontaneously arrange into bilayers with hydrophobic tails pointing inward and hydrophilic heads facing outward, forming the basic structure of all cell membranes.

Worked Example

A triglyceride is formed from glycerol (${\text{C}}_3{\text{H}}_8{\text{O}}_3$) and three identical stearic acid molecules (${\text{C}}_{17}{\text{H}}_{35}\text{COOH}$). Write the balanced equation for this condensation reaction: $${\text{C}}_3{\text{H}}_8{\text{O}}_3+3{\text{C}}_{18}{\text{H}}_{36}{\text{O}}_2\to {\text{C}}_{57}{\text{H}}_{110}{\text{O}}_6+3{\text{H}}_2\text{O}$$ Mark schemes award marks for correct atom counts and identification of water as a product of the condensation reaction.

5. Proteins — primary to quaternary structure

Proteins are polymers of amino acid monomers, joined by peptide bonds formed via condensation reactions. All amino acids have the same core structure: a central carbon bonded to an amine group ($-{\text{NH}}_2$), a carboxylic acid group ($-\text{COOH}$), a hydrogen atom, and a variable R group (20 different R groups exist for the 20 common biologically active amino acids). Protein structure is organised into four hierarchical levels:

1. Primary structure: The linear sequence of amino acids in the polypeptide chain, determined directly by the base sequence of the gene that codes for the protein. Held together only by peptide bonds. A single amino acid mutation can completely alter protein function.
2. Secondary structure: Regular local folding of the polypeptide chain into either an $\alpha$-helix (coiled shape) or $\beta$-pleated sheet (flat folded shape), held together by hydrogen bonds between the $-\text{NH}$ of one peptide bond and the $-\text{C=O}$ of an adjacent peptide bond.
3. Tertiary structure: The overall 3D globular shape of a single folded polypeptide chain, stabilised by four types of bonds between R groups: hydrogen bonds, ionic bonds between charged R groups, disulfide bridges between cysteine R groups, and hydrophobic interactions between non-polar R groups. The tertiary structure directly determines protein function, e.g., the shape of an enzyme’s active site.
4. Quaternary structure: The structure of proteins made of two or more polypeptide subunits, held together by the same non-peptide bonds found in tertiary structure. For example, haemoglobin has four subunits (two $\alpha$ chains, two $\beta$ chains) each with an iron-containing haem group that binds oxygen.

Worked Example

A polypeptide chain has 247 amino acids. How many peptide bonds does it contain? For a linear polypeptide with $n$ amino acids, the number of peptide bonds is $n-1$, so $247-1=246$ peptide bonds.

6. Enzymes intro — induced fit, active site

Enzymes are globular proteins that act as biological catalysts, lowering the activation energy of metabolic reactions without being used up or altered in the reaction.

• The active site is a specific cleft or pocket on the enzyme surface with a shape and chemical environment complementary to its specific substrate molecule. Only the correct substrate can bind to the active site, giving enzymes their high specificity.
• The original lock and key model proposed the active site was a rigid, fixed shape perfectly complementary to the substrate. This has been replaced by the induced fit model, the accepted explanation for enzyme action: when the substrate binds to the active site, the active site changes shape slightly to fit the substrate more tightly. This shape change puts strain on the substrate’s chemical bonds, lowering activation energy far more effectively than a rigid active site. Evidence for this model comes from X-ray crystallography images that show measurable shape changes in enzymes after substrate binding.

Worked Example

A mutation changes one amino acid in the active site of the enzyme sucrase, which breaks down sucrose. Explain why the mutated enzyme can no longer catalyse sucrose breakdown. The amino acid change alters the R group interactions that stabilise the tertiary structure of the enzyme, changing the 3D shape of the active site. The active site is no longer complementary to the shape of sucrose, so the substrate cannot bind to form enzyme-substrate complexes, and no reaction is catalysed.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Drawing the hydroxyl group on carbon 1 of $\alpha$-glucose above the ring, mixing it up with $\beta$-glucose. Why students do it: They fail to memorise the small structural difference between the two isomers. Correct move: Remember the mnemonic "α = below, β = above" for the C1 hydroxyl group. Mark schemes explicitly award marks for correct orientation, so always label carbon 1 when drawing glucose isomers.
• Pitfall 2: Stating that triglycerides are polymers. Why students do it: They assume all large biological molecules are polymers. Correct move: Polymers are made of repeating identical monomer units; triglycerides have no repeating monomer, so they are not classified as polymers. Only polysaccharides, proteins, and nucleic acids are considered polymers in this topic.
• Pitfall 3: Only mentioning hydrogen bonds when explaining tertiary protein structure. Why students do it: They forget the other stabilising bond types. Correct move: List at least three bond types (hydrogen, ionic, disulfide, hydrophobic interactions) when describing tertiary structure to gain full marks; examiners regularly require multiple bond types for 2+ mark questions.
• Pitfall 4: Confusing water’s high specific heat capacity with its high latent heat of vaporisation when explaining cooling via sweating. Why students do it: They mix up the two thermal properties. Correct move: SHC = temperature stability; latent heat of vaporisation = cooling effect when water evaporates. Explicitly name the correct property for each biological role.
• Pitfall 5: Describing the induced fit model as identical to the lock and key model. Why students do it: They oversimplify enzyme action to just "complementary shape". Correct move: Always mention that the active site changes shape after substrate binding when explaining induced fit; mark schemes deduct marks if you omit this key detail.

8. Practice Questions (A-Level Biology Style)

Question 1

(a) Explain how the polarity of water molecules allows it to act as a solvent for sodium chloride (NaCl) in blood plasma. [3 marks] (b) Calculate the amount of energy required to raise the temperature of 1.5 kg of water by 3°C, using $c=4200{\text{ J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}$. [2 marks]

Solution

(a) 1. Water is a polar molecule with ${\delta }^{-}$ oxygen and ${\delta }^{+}$ hydrogen atoms [1]. 2. NaCl dissociates into ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions when added to water [1]. 3. ${\delta }^{+}$ hydrogen atoms surround ${\text{Cl}}^{-}$ ions, and ${\delta }^{-}$ oxygen atoms surround ${\text{Na}}^{+}$ ions, forming hydration shells that keep the ions dissolved in plasma [1]. (b) Use $Q=mc\Delta T$: $$Q=1.5\times 4200\times 3=18900\text{ J}=18.9\text{ kJ}$$ [1 mark for correct substitution, 1 mark for final answer in correct units]

Question 2

Compare the structure and function of phospholipids and triglycerides. [4 marks]

Solution

Structure: Both have a glycerol backbone attached to fatty acids via ester bonds [1]. Triglycerides have 3 fatty acid tails, while phospholipids have 2 fatty acid tails and one charged phosphate head group [1]. Function: Triglycerides are used for long-term energy storage, thermal insulation, and buoyancy [1]. Phospholipids are amphipathic, so they form bilayers that make up the structure of all cell membranes [1].

Question 3

Explain how a change in the primary structure of an enzyme can lead to a non-functional protein, even if the mutation is not in the active site. [3 marks]

Solution

1. The primary structure is the sequence of amino acids, which determines all higher levels of protein structure [1]. 2. A mutation that changes an amino acid outside the active site can alter R group interactions that stabilise the tertiary structure of the enzyme [1]. 3. This changes the overall 3D shape of the protein, including the shape of the active site, so the substrate can no longer bind, making the enzyme non-functional [1].

9. Quick Reference Cheatsheet

10. What's Next

This topic is the foundation for almost every subsequent unit in the A-Level Biology syllabus. You will apply your knowledge of phospholipid bilayers to cell membrane structure and transport in Topic 4, protein structure to enzyme inhibition and immunology in later units, and carbohydrate structure to plant transport and cell wall function in Topic 7. Biological molecules content also forms the basis of key practical assessments in Papers 3 and 5, including qualitative food tests for reducing sugars, starch, proteins, and lipids, so mastering this content now will reduce your practical revision workload later.

If you are stuck on any concept, from drawing glycosidic or ester bonds to explaining the difference between the lock and key and induced fit models, you can ask Ollie for step-by-step explanations, custom flashcards, or additional practice questions tailored to your weak spots at any time by visiting the homepage. Once you have mastered the content in this guide, you can move on to our next A-Level Biology study guide on Enzyme Kinetics, which builds on your basic knowledge of enzyme action to cover factors affecting reaction rate and enzyme inhibition.

Aligned with the Cambridge International AS & A Level Biology 9700 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.