Biological Molecules

All living organisms are composed of a relatively small number of chemical elements. Carbon, hydrogen, oxygen, nitrogen, phosphorus, and sulphur form the backbone of the four main groups of biological molecules: carbohydrates, lipids, proteins, and nucleic acids. In addition, inorganic ions play essential roles in biological processes. Understanding their structure, function, and the tests used to identify them is the foundation of A-Level Biology.

Key Definition: A monomer is a small, repeating molecular unit that can be joined together to form a polymer. A polymer is a large molecule made of many monomers joined by condensation reactions.

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Water — The Solvent of Life

Before examining the macromolecules, it is essential to appreciate the unique properties of water. Water is a polar molecule: the oxygen atom is slightly electronegatively charged (δ−) and the hydrogen atoms are slightly positively charged (δ+). This polarity allows water molecules to form hydrogen bonds with one another and with other polar molecules.

Property	Explanation	Biological Importance
High specific heat capacity	Hydrogen bonds absorb a large amount of energy before temperature rises	Maintains stable body temperature and aquatic environments
Cohesion and surface tension	Hydrogen bonds hold water molecules together	Allows transpiration pull in xylem; insects walk on pond surfaces
Excellent solvent	Polar molecules dissolve readily	Enables transport of ions and metabolites in blood and cytoplasm
Maximum density at 4 °C	Ice is less dense than liquid water	Lakes freeze from the top, insulating organisms beneath
High latent heat of vaporisation	Considerable energy needed to evaporate water	Sweating and transpiration provide effective cooling

Exam Tip: When asked to explain a property of water, always link the explanation back to hydrogen bonding between water molecules. Simply stating "water is a good solvent" without mentioning polarity and hydrogen bonds will not gain full marks.

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Carbohydrates

Carbohydrates contain carbon, hydrogen, and oxygen in the ratio C_n(H₂O)_n.

Monosaccharides

Key Definition: A monosaccharide is the simplest unit of a carbohydrate; a single sugar molecule that cannot be hydrolysed further.

• Glucose (C₆H₁₂O₆) — the primary respiratory substrate. Exists as α-glucose and β-glucose, differing only in the position of the hydroxyl group on carbon-1. In α-glucose the –OH on C1 is below the plane of the ring; in β-glucose it is above.
• Galactose and fructose — structural isomers of glucose.
• Ribose (C₅H₁₀O₅) — a pentose sugar found in RNA and ATP.
• Deoxyribose (C₅H₁₀O₄) — a pentose sugar found in DNA (one fewer oxygen atom than ribose).

A diagram of α-glucose would show a hexagonal ring with carbons numbered 1–5 around the ring and carbon 6 projecting upward from C5. The hydroxyl (–OH) group on C1 points downward, while in β-glucose it points upward. This seemingly small difference has enormous consequences for the polymers they form.

Disaccharides

Formed by condensation reactions (glycosidic bonds) between two monosaccharides, releasing water:

Disaccharide	Monomers	Bond
Maltose	α-glucose + α-glucose	α-1,4 glycosidic
Sucrose	α-glucose + fructose	α-1,2 glycosidic
Lactose	β-galactose + α-glucose	β-1,4 glycosidic

Key Definition: A glycosidic bond is a covalent bond formed between two monosaccharides by a condensation reaction, with the removal of water.

Disaccharides are broken down into monosaccharides by hydrolysis — the addition of water to break the glycosidic bond, catalysed by specific enzymes (e.g., maltase, sucrase, lactase).

Polysaccharides

• Starch (amylose + amylopectin) — energy storage in plants; compact, insoluble, does not affect water potential. Amylose is an unbranched helix of α-glucose units linked by α-1,4 glycosidic bonds. Amylopectin has α-1,4 bonds with α-1,6 branches, allowing faster hydrolysis at the many terminal ends.
• Glycogen — energy storage in animals; very highly branched (more than amylopectin) for rapid hydrolysis when glucose is needed quickly, e.g., during exercise. Stored mainly in liver and muscle cells.
• Cellulose — structural polysaccharide in plant cell walls; β-1,4 glycosidic bonds create straight chains. Alternate glucose molecules are rotated 180°, allowing extensive hydrogen bonding between parallel chains to form microfibrils. These microfibrils provide enormous tensile strength, preventing plant cells from bursting when turgid.

Exam Tip: A very common exam question asks you to compare starch and cellulose. Remember: both are made of glucose, but starch uses α-glucose (coiled, compact, storage) and cellulose uses β-glucose (straight chains, structural). The bond orientation is the key difference.

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Lipids

Lipids are non-polar, hydrophobic molecules. The main types at A-Level are triglycerides and phospholipids.

Triglycerides

Key Definition: A triglyceride is a lipid molecule formed from one molecule of glycerol and three fatty acid molecules joined by ester bonds through condensation reactions.

• One glycerol + three fatty acids joined by ester bonds via condensation.
• Saturated fatty acids have no C=C double bonds (straight chains, solid at room temperature).
• Unsaturated fatty acids contain one or more C=C double bonds (kinked chains, liquid at room temperature). Monounsaturated = one double bond; polyunsaturated = two or more double bonds.
• Functions: long-term energy storage (more energy per gram than carbohydrates because of the higher ratio of hydrogen to oxygen atoms), thermal insulation, protection of organs, waterproofing, electrical insulation in myelin sheaths.

Phospholipids

• Two fatty acid tails (hydrophobic) + one phosphate head (hydrophilic) → amphipathic.
• Spontaneously form bilayers in aqueous environments — the basis of all cell membranes.
• A diagram of a phospholipid would show a circular head (representing the hydrophilic phosphate group) attached to two wavy parallel lines (representing the hydrophobic fatty acid tails).

Cholesterol

• A type of lipid with a four-ring hydrocarbon structure.
• Fits between phospholipids in cell membranes, regulating fluidity: at high temperatures it reduces fluidity by restricting phospholipid movement; at low temperatures it prevents the membrane from becoming too rigid.

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Proteins

Proteins are polymers of amino acids linked by peptide bonds.

Amino Acid Structure

Key Definition: An amino acid is a monomer of a protein, consisting of a central carbon atom bonded to an amino group (–NH₂), a carboxyl group (–COOH), a hydrogen atom, and a variable R group (side chain).

Every amino acid has a central carbon bonded to an amino group (–NH₂), a carboxyl group (–COOH), a hydrogen atom, and a variable R group (side chain). It is the R group that differs between the 20 naturally occurring amino acids, giving each its unique chemical properties.

Levels of Protein Structure

Level	Description	Bonds/Interactions Involved
Primary	Linear sequence of amino acids in the polypeptide chain	Peptide bonds
Secondary	Local folding into α-helices or β-pleated sheets	Hydrogen bonds between C=O and N–H groups of the peptide backbone
Tertiary	Overall 3D shape of a single polypeptide	Ionic bonds, disulphide bridges, hydrophobic interactions, hydrogen bonds between R groups
Quaternary	Association of two or more polypeptide subunits, sometimes with prosthetic groups	Same interactions as tertiary, plus bonds between subunits

R-group interactions in tertiary structure (detail):

• Ionic bonds — form between positively charged (e.g., lysine) and negatively charged (e.g., aspartate) R groups. Broken by changes in pH.
• Disulphide bridges — strong covalent bonds between the sulphur atoms of two cysteine residues. Not easily broken.
• Hydrophobic interactions — non-polar R groups cluster in the interior of the protein, away from water.
• Hydrogen bonds — weak individually but numerous, forming between polar R groups.

Proteins can be globular (soluble, metabolic roles — enzymes, antibodies, haemoglobin) or fibrous (insoluble, structural roles — collagen, keratin). Globular proteins fold into compact spherical shapes with hydrophilic R groups on the outside (making them soluble). Fibrous proteins have long, repetitive sequences that form regular structures such as the triple helix of collagen.

Chromatography for Amino Acid Separation

Amino acids in a protein can be identified using thin-layer chromatography or paper chromatography:

1. The protein is first hydrolysed into its constituent amino acids.
2. A small spot of the amino acid mixture is placed on the origin line of chromatography paper.
3. The paper is placed in a suitable solvent.
4. As the solvent moves up the paper by capillary action, different amino acids are carried different distances depending on their solubility in the solvent.
5. The paper is dried and sprayed with ninhydrin to reveal the positions of the amino acids as coloured spots.
6. Each amino acid is identified by calculating its Rf value: Rf = distance moved by amino acid ÷ distance moved by solvent front.

Exam Tip: Rf values are always between 0 and 1. If the question provides a table of known Rf values, match the calculated Rf value to identify the amino acid. Always measure from the origin line, not from the bottom of the paper.

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Enzymes

Key Definition: An enzyme is a biological catalyst — a protein (usually) that speeds up the rate of a metabolic reaction by lowering the activation energy, without being used up in the reaction.

Enzyme-Substrate Complex

Enzymes work by forming a temporary enzyme-substrate complex at the active site — a specific region on the enzyme surface with a shape complementary to the substrate.

Lock-and-Key Model

• Proposed by Emil Fischer (1894).
• The active site has a fixed shape that is exactly complementary to the substrate, like a key fitting a lock.
• This model explains enzyme specificity but does not account for changes in active site shape during catalysis.

Induced Fit Model

• Proposed by Daniel Koshland (1958).
• The active site has a flexible shape that changes slightly when the substrate binds, moulding around the substrate to form the enzyme-substrate complex.
• This conformational change places strain on the substrate, lowering the activation energy.
• The induced fit model is now the accepted explanation and better accounts for experimental observations.

A diagram of the induced fit model would show the enzyme active site as a slightly open cleft. When the substrate approaches, the active site changes shape to wrap closely around the substrate. After the reaction, the products are released and the active site returns to its original shape.

Factors Affecting Enzyme Activity

Temperature:

• Increasing temperature increases the kinetic energy of molecules, so enzyme-substrate collisions occur more frequently and with more energy → rate increases.
• Beyond the optimum temperature (around 37 °C for human enzymes), the enzyme begins to denature: hydrogen bonds and ionic bonds maintaining the tertiary structure are disrupted, the active site changes shape, and the substrate can no longer bind.
• Denaturation is usually permanent for most enzymes.

pH:

• Each enzyme has an optimum pH (e.g., pepsin ≈ pH 2; salivary amylase ≈ pH 7; trypsin ≈ pH 8).
• Extreme pH values alter the ionisation of R groups, breaking ionic bonds and hydrogen bonds, causing the active site to change shape → denaturation.

Substrate concentration:

• Increasing substrate concentration increases the rate of reaction until all active sites are occupied (enzyme saturation).
• At saturation, the reaction rate reaches Vmax (maximum velocity) and further increases in substrate concentration have no effect.

Enzyme concentration:

• More enzyme molecules = more active sites available = faster rate (provided substrate is in excess).

Enzyme Kinetics: Vmax and Km

Key Definition: Vmax is the maximum rate of reaction when all enzyme active sites are saturated with substrate. Km (Michaelis constant) is the substrate concentration at which the reaction rate is half of Vmax; it is a measure of the enzyme's affinity for its substrate.

• A low Km indicates high affinity — the enzyme reaches half-Vmax at a low substrate concentration.
• A high Km indicates low affinity — a higher substrate concentration is needed to reach half-Vmax.

Enzyme Inhibition

Type	Mechanism	Effect on Vmax	Effect on Km
Competitive	Inhibitor has a similar shape to the substrate and competes for the active site	Unchanged (can be overcome by increasing substrate concentration)	Increased (apparent lower affinity)
Non-competitive	Inhibitor binds to an allosteric site (not the active site), causing the active site to change shape	Decreased (cannot be overcome by increasing substrate concentration)	Unchanged

Exam Tip: A classic exam question shows a graph of rate vs substrate concentration with and without an inhibitor. For competitive inhibition, Vmax is the same but the curve is shifted to the right. For non-competitive inhibition, Vmax is lower but the curve starts at the same point. Practise sketching both.

Worked Example 1 — Enzyme calculations:

An enzyme-catalysed reaction produces 0.15 µmol of product per minute at a substrate concentration of 2.0 mmol/L. The Vmax for this enzyme is 0.30 µmol/min. Calculate the Km.

Solution: Using the relationship: at [S] = Km, rate = Vmax / 2

Rate = 0.15 µmol/min = 0.30 / 2 = Vmax / 2

Therefore, Km = 2.0 mmol/L (because this substrate concentration gives exactly half the maximum rate).

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Nucleic Acids

DNA and RNA are polynucleotides composed of nucleotide monomers.

Key Definition: A nucleotide consists of a pentose sugar, a phosphate group, and a nitrogenous base. Nucleotides are joined by phosphodiester bonds to form polynucleotides.

• Each nucleotide = pentose sugar + phosphate group + nitrogenous base.
• DNA: deoxyribose sugar, bases A, T, C, G; double-stranded helix with antiparallel strands joined by complementary base pairing (A=T with 2 hydrogen bonds, C≡G with 3 hydrogen bonds).
• RNA: ribose sugar, bases A, U, C, G; usually single-stranded.
• ATP (adenosine triphosphate) is a modified nucleotide that acts as the universal energy currency of cells. It is hydrolysed to ADP + Pi, releasing approximately 30.6 kJ/mol of energy.

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Inorganic Ions

Inorganic ions, though required in small amounts, are essential for many biological processes:

Ion	Role
Iron (Fe²⁺)	Component of the haem group in haemoglobin; binds reversibly to oxygen for transport. Also a cofactor in cytochrome oxidase in the electron transport chain.
Phosphate (PO₄³⁻)	Component of ATP, nucleotides, and phospholipids. Essential for phosphorylation reactions that activate enzymes and for the sugar-phosphate backbone of DNA/RNA.
Calcium (Ca²⁺)	Essential for muscle contraction (binds to troponin, exposing binding sites on actin). Required for blood clotting and as a structural component of bones and teeth. Acts as a second messenger in cell signalling.
Hydrogen (H⁺)	Determines pH; critical in chemiosmosis (proton gradient across membranes drives ATP synthesis).

Exam Tip: Questions on inorganic ions often appear in unexpected contexts. If asked about the role of iron, do not just say "in haemoglobin" — state that Fe²⁺ is part of the haem prosthetic group and that it reversibly binds to oxygen.

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Biochemical Tests

Test	Reagent	Positive Result
Reducing sugars	Benedict's solution (heat to ~80 °C in water bath)	Green → yellow → orange → brick-red precipitate
Non-reducing sugars	Acid hydrolysis (boil with HCl), neutralise with NaHCO₃, then Benedict's	Colour change after hydrolysis
Starch	Iodine solution (in KI)	Blue-black colour
Proteins	Biuret reagent (NaOH + CuSO₄)	Purple/lilac colour
Lipids	Ethanol emulsion test	White emulsion

Benedict's Test as a Semi-Quantitative Method

The Benedict's test can be made semi-quantitative to estimate the concentration of reducing sugar in a sample:

Method using colorimetry:

1. Prepare a series of serial dilutions of a known reducing sugar (e.g., glucose) at concentrations such as 0.0, 0.2, 0.4, 0.6, 0.8, and 1.0 mol/dm³.
2. Add equal volumes of Benedict's reagent to each solution and heat in a water bath at 80 °C for a set time (e.g., 5 minutes).
3. Filter each sample to remove the precipitate.
4. Use a colorimeter (set to a blue filter, ~440 nm) to measure the absorbance or transmission of each filtrate.
5. Plot a calibration curve of absorbance against known sugar concentration.
6. Test the unknown sample under identical conditions and use the calibration curve to determine its sugar concentration.

Worked Example 2 — Serial dilution calculation:

You have a 2.0 mol/dm³ glucose solution. Describe how to make 10 cm³ of a 0.5 mol/dm³ solution.

Solution: Using C₁V₁ = C₂V₂: 2.0 × V₁ = 0.5 × 10 V₁ = 5.0 / 2.0 = 2.5 cm³

Take 2.5 cm³ of the 2.0 mol/dm³ stock solution and add 7.5 cm³ of distilled water to make a total volume of 10 cm³ at 0.5 mol/dm³.

Worked Example 3 — Magnification calculation (relevant to biochemistry microscopy of starch grains):

A starch grain has an actual diameter of 25 µm. It appears 50 mm across in a photomicrograph. What is the magnification?

Solution: First, convert to the same units: 50 mm = 50,000 µm. Magnification = image size ÷ actual size = 50,000 ÷ 25 = ×2000.

Exam Tip: Always convert units before calculating magnification. The most common error is mixing mm and µm. Remember: 1 mm = 1000 µm.