Chemistry of Life

Elements and Compounds (CED Unit 1)

Essential Elements

The most common elements in living organisms:

Element	Symbol	Role
Carbon	C	Backbone of organic molecules
Hydrogen	H	Component of water, organic molecules
Oxygen	O	Component of water, electron acceptor
Nitrogen	N	Amino acids, nucleotides
Phosphorus	P	ATP, nucleic acids, phospholipids
Sulfur	S	Amino acids (cysteine, methionine)

Trace elements (required in small amounts): Fe, Cu, Zn, Mn, I, Mo.

Worked Example: Why specific trace elements matter. Iron (Fe) is a component of haemoglobin, the Oxygen-carrying protein in red blood cells. Without iron, haemoglobin cannot be synthesised, leading To iron-deficiency anaemia. Iodine (I) is required for the synthesis of thyroxine, a hormone that Regulates metabolism. Iodine deficiency causes goitre (enlarged thyroid gland).

Bonding in Biological Molecules

Covalent bonds: Strong, stable bonds sharing electrons (C—C, C—H, C—O, C—N).
Hydrogen bonds: Weak, between H bonded to N/O/F and a lone pair on N/O/F. Critical for water properties, protein secondary structure, and DNA base pairing.
Ionic bonds: Between charged groups (e.g., in salt bridges in proteins).
Van der Waals forces: Weak, transient attractions between molecules.

Comparison of bond strengths in biological systems:

Bond Type	Strength (kJ/mol)	Biological Role
Covalent	200—500	Backbone of organic molecules
Ionic	100—400	Salt bridges in proteins, bone mineral structure
Hydrogen	5—40	DNA base pairing, protein secondary structure, water
Van der Waals	0.5—5	Protein folding, lipid interactions

Water (CED Unit 1)

Water is essential for life due to its unique properties:

Cohesion and adhesion: H-bonds between water molecules (cohesion) and between water and other surfaces (adhesion). Enables capillary action and transpiration in plants.
High specific heat: $4.18 \mathrm{ J/(g\cdot^\circ\mathrm{C)$ . Moderates temperature changes.
High heat of vaporization: Requires significant energy to evaporate. Provides cooling through sweating.
Lower density as a solid: Ice floats, insulating aquatic environments.
Universal solvent: Polar molecules and ions dissolve .
High surface tension: Due to cohesion at the air-water interface.

Worked Example: How water properties support life.

Cohesion: Water molecules stick together due to hydrogen bonding. This allows water to be pulled up the xylem in plants (transpiration stream) as a continuous column.
High specific heat: Oceans absorb and release large amounts of heat with minimal temperature change, creating stable climates near coastlines.
Ice floats: In winter, ice forms on the surface of lakes, insulating the water below and allowing aquatic organisms to survive.
Universal solvent: Many biochemical reactions occur in aqueous solution because water dissolves ions, sugars, amino acids, and other polar molecules.

pH and Buffers

\mathrm{pH = -\log[\mathrm{H^+]

pH	Classification
$\lt 7$	Acidic
$7$	Neutral
$\gt 7$	Basic

Buffers resist pH changes. The carbonic acid/bicarbonate buffer system in blood:

\mathrm{H_2\mathrm{CO_3 \rightleftharpoons \mathrm{H^+ + \mathrm{HCO_3^-

When $[\mathrm{H^+]$ increases, the equilibrium shifts left, consuming excess $\mathrm{H^+$ .

Worked Example: pH calculations.

A solution has $[\mathrm{H^+] = 1 \times 10^{-4}$ M.

$\mathrm{pH = -\log(1 \times 10^{-4}) = 4$

If the solution is diluted by a factor of 10, $[\mathrm{H^+] = 1 \times 10^{-5}$ M, and $\mathrm{pH = 5$ .

A one-unit increase in pH represents a tenfold decrease in $[\mathrm{H^+]$ .

Worked Example: Buffer action.

During vigorous exercise, lactic acid is produced, releasing $\mathrm{H^+$ into the blood. Without The buffer system, blood pH would drop dangerously.

The bicarbonate buffer system responds:

$\mathrm{H^+ + \mathrm{HCO_3^- \to \mathrm{H_2\mathrm{CO_3 \to \mathrm{H_2\mathrm{O + \mathrm{CO_2$

The excess $\mathrm{H^+$ is consumed by reacting with $\mathrm{HCO_3^-$ . The resulting $\mathrm{CO_2$ is removed by increased breathing rate. Blood pH is maintained close to 7.4.

Macromolecules (CED Unit 1)

Carbohydrates

General formula: $\mathrm{C_n(\mathrm{H_2\mathrm{O)_n$

Monomers: Monosaccharides (glucose, fructose, galactose, ribose, deoxyribose).

Polymerization: Dehydration synthesis (condensation) forms glycosidic bonds. Hydrolysis breaks Them.

Type	Structure	Function	Examples
Monosaccharide	Single sugar ring	Energy source, building block	Glucose, fructose
Disaccharide	Two monosaccharides	Energy source	Sucrose, lactose, maltose
Polysaccharide	Long chain of sugars	Energy storage, structural support	Starch, glycogen, cellulose

Key polysaccharides:

Starch: Energy storage in plants. Amylose (unbranched) + amylopectin (branched). $\alpha$ -glycosidic linkages (digestible).
Glycogen: Energy storage in animals. Highly branched. $\alpha$ -glycosidic linkages.
Cellulose: Structural in plant cell walls. Unbranched. $\beta$ -glycosidic linkages (indigestible by most animals).
Chitin: Structural in arthropod exoskeletons and fungal cell walls. Modified glucose with nitrogen-containing groups.

Worked Example: Alpha vs. Beta linkages.

In starch, glucose monomers are joined by $\alpha$ -1,4-glycosidic bonds. The glucose rings are in The $\alpha$ configuration (OH group below the ring at C1). This produces a helical structure that Can be coiled, making starch compact for storage. Human amylase can hydrolyse $\alpha$ -glycosidic Bonds, so starch is digestible.

In cellulose, glucose monomers are joined by $\beta$ -1,4-glycosidic bonds. The glucose rings are in The $\beta$ configuration (OH group above the ring at C1). Every other glucose is flipped 180 Degrees, producing straight chains that form hydrogen bonds with neighbouring chains, creating Strong, rigid fibres. Human amylase cannot hydrolyse $\beta$ -glycosidic bonds, so cellulose is Indigestible by humans.

Lipids

Lipids are not true polymers but are hydrophobic organic molecules.

Type	Structure	Function
Triglycerides	Glycerol + 3 fatty acids	Energy storage, insulation
Phospholipids	Glycerol + 2 fatty acids + phosphate	Cell membrane structure
Steroids	Four fused rings	Hormones, cholesterol
Waxes	Long-chain alcohol + fatty acid	Waterproof coatings

Saturated vs unsaturated fatty acids:

Saturated: No double bonds, straight chains, solid at room temperature.
Unsaturated: One or more double bonds, kinked chains, liquid at room temperature.
Trans fats: Unsaturated fats with hydrogens on opposite sides of double bonds. Associated with cardiovascular disease.

Phospholipids: Amphipathic — hydrophilic phosphate head and hydrophobic fatty acid tails. Form The bilayer of cell membranes.

Worked Example: Structure of a triglyceride.

A triglyceride consists of one glycerol molecule (a 3-carbon alcohol with three OH groups) joined to Three fatty acid molecules (long hydrocarbon chains with a carboxyl group at one end) by three ester Bonds. Each ester bond is formed by a condensation reaction, releasing one water molecule.

$\mathrm{Glycerol + 3 \mathrm{ Fatty acids \to \mathrm{Triglyceride + 3 \mathrm{ H_2\mathrm{O$

The long hydrocarbon tails are hydrophobic, making triglycerides insoluble in water. This makes them Ideal for energy storage because they do not affect the water potential of cells.

Proteins

Monomers: Amino acids (20 common types).

Structure of an amino acid: Central carbon ( $\alpha$ -carbon) bonded to:

Amino group ( $\mathrm{NH_2$ )
Carboxyl group ( $\mathrm{COOH$ )
Hydrogen atom
R group (side chain, varies)

Polymerization: Peptide bonds (amide bonds) form by dehydration synthesis between the carboxyl Group of one amino acid and the amino group of another.

Four levels of protein structure:

Primary: Linear sequence of amino acids. Determined by the gene.
Secondary: Local folding patterns stabilized by hydrogen bonds.

$\alpha$ -helix: right-handed coiled structure.
$\beta$ -pleated sheet: extended, zigzag structure.

Tertiary: Overall 3D shape of a single polypeptide. Stabilized by hydrogen bonds, disulfide bridges, ionic interactions, and hydrophobic interactions.
Quaternary: Assembly of multiple polypeptide subunits (e.g., hemoglobin has 4 subunits).

Denaturation: Loss of protein structure (secondary, tertiary, quaternary) due to heat, pH Change, or chemical disruption. The primary structure remains intact.

Worked Example: How disulfide bridges stabilize protein structure.

Disulfide bridges (also called disulfide bonds) form between the sulfur atoms of two cysteine Residues. These are covalent bonds, which are much stronger than hydrogen bonds or ionic Interactions. In proteins like antibodies, disulfide bridges hold the polypeptide chains together, Maintaining the Y-shaped structure even under stress. Reducing agents that break disulfide bridges Can cause the protein to unfold.

Nucleic Acids

Monomers: Nucleotides, each consisting of:

A 5-carbon sugar (ribose or deoxyribose)
A phosphate group
A nitrogenous base

Nitrogenous bases:

Purines: Adenine (A), Guanine (G) — double ring
Pyrimidines: Cytosine (C), Thymine (T, DNA only), Uracil (U, RNA only) — single ring

DNA:

Double helix (Watson and Crick, 1953)
Antiparallel strands ( $5' \to 3'$ and $3' \to 5'$ )
Sugar-phosphate backbone on the outside, bases on the inside
Base pairing: A—T (2 H-bonds), G—C (3 H-bonds)
Chargaff’s rules: $[\mathrm{A] = [\mathrm{T]$ , $[\mathrm{G] = [\mathrm{C]$

RNA:

single-stranded
Contains uracil instead of thymine
Sugar is ribose (not deoxyribose)
Types: mRNA (messenger), tRNA (transfer), rRNA (ribosomal)

Worked Example: Chargaff’s rules applied.

If a DNA molecule has 20% adenine, then by Chargaff’s rules, it also has 20% thymine. Since A + T + G + C = 100%, we have G + C = 60%. If G = C (Chargaff’s rules), then G = 30% and C = 30%.

A DNA molecule with 30% GC content would have 30% G, 30% C, 20% A, and 20% T.

Enzymes (CED Unit 1)

Properties

Biological catalysts: Speed up reactions without being consumed.
Protein (most enzymes) or RNA (ribozymes).
Highly specific for substrates (lock-and-key or induced-fit model).
Lower activation energy ( $E_a$ ) but do not change $\Delta G$ .

Enzyme-Substrate Complex

E + S \rightleftharpoons ES \rightleftharpoons E + P

Factors Affecting Enzyme Activity

Temperature: Rate increases until the optimal temperature, then denaturation occurs.
pH: Each enzyme has an optimal pH. Pepsin (stomach) $\approx 2$ ; trypsin (small intestine) $\approx 8$ .
Substrate concentration: Rate increases with [S] until all active sites are saturated ( $V_{\max}$ is reached). Described by the Michaelis-Menten equation.
Enzyme concentration: Rate is proportional to [E] when substrate is not limiting.

Michaelis-Menten Kinetics

V = \frac{V_{\max}[S]}{K_m + [S]}

$V_{\max}$ : maximum rate when enzyme is saturated.
$K_m$ : substrate concentration at half $V_{\max}$ . Lower $K_m$ means higher affinity.

Worked Example: Calculating $V_{\max}$ and $K_m$ .

An enzyme has the following rates at different substrate concentrations:

[S] ( $\mu$ M)	v ( $\mu$ Mol/min)
2	5.0
5	10.0
10	14.3
20	16.7
50	18.9
100	19.6

At very high [S], $v$ approaches $V_{\max} \approx 20$ $\mu$ Mol/min.

At $v = V_{\max}/2 = 10$ [S] = 5 $\mu$ M, so $K_m = 5$ $\mu$ M.

This means the enzyme reaches half its maximum rate when the substrate concentration is 5 $\mu$ M. A Lower $K_m$ would indicate a higher affinity between enzyme and substrate.

Enzyme Inhibition

Type	Binding Site	Reversibility	Effect on $K_m$	Effect on $V_{\max}$
Competitive	Active site	Reversible	Increases	No change
Noncompetitive	Allosteric site	Reversible	No change	Decreases
Uncompetitive	ES complex only	Reversible	Decreases	Decreases
Irreversible	Active or allosteric	Irreversible	—	Decreases

Worked Example: Why competitive inhibition increases $K_m$ but not $V_{\max}$ .

A competitive inhibitor binds to the active site, competing directly with the substrate. At low Substrate concentration, the inhibitor effectively blocks many active sites, reducing the reaction Rate. To achieve half $V_{\max}$ A higher substrate concentration is needed (to outcompete the Inhibitor), so the apparent $K_m$ increases.

However, at very high substrate concentrations, the substrate outcompetes the inhibitor for all Active sites, and the enzyme reaches the same $V_{\max}$ as without the inhibitor. Therefore, $V_{\max}$ is unchanged.

A noncompetitive inhibitor binds to an allosteric site (not the active site), changing the enzyme’s Shape so that it works less efficiently. Even at very high substrate concentrations, some enzymes Remain in the less efficient form, so $V_{\max}$ decreases. Since the active site is not directly Affected, the enzyme’s affinity for the substrate ( $K_m$ ) does not change.

Free Energy and Metabolism (CED Unit 1)

Endergonic vs Exergonic Reactions

\Delta G = \Delta H - T\Delta S

Type	$\Delta G$	Description
Exergonic	$\lt 0$	Spontaneous, releases energy
Endergonic	$\gt 0$	Nonspontaneous, requires energy input

ATP: The Energy Currency

ATP (adenosine triphosphate) stores energy in its high-energy phosphate bonds.

\mathrm{ATP \rightleftharpoons \mathrm{ADP + \mathrm{P_i + \mathrm{energy \quad (\Delta G \approx -30.5 \mathrm{ kJ/mol)

ATP hydrolysis is exergonic and couples with endergonic reactions to drive them.

Coupled Reactions

The energy released from ATP hydrolysis drives unfavorable reactions by coupling:

\mathrm{Overall: \mathrm{ATP + \mathrm{Glucose \to \mathrm{ADP + \mathrm{Glucose-6-phosphate \quad (\Delta G \lt 0)

Worked Example: Energy coupling in cells.

The phosphorylation of glucose to glucose-6-phosphate has $\Delta G \approx +13.8$ kJ/mol (endergonic). ATP hydrolysis has $\Delta G \approx -30.5$ kJ/mol (exergonic). When coupled:

$\Delta G_{\mathrm{overall} = +13.8 + (-30.5) = -16.7 \mathrm{ kJ/mol$

The overall reaction is now exergonic (spontaneous), so the cell can phosphorylate glucose using the Energy from ATP hydrolysis.

Common Pitfalls

Confusing dehydration synthesis with hydrolysis. Dehydration removes water to build polymers; hydrolysis adds water to break them.
Confusing $\alpha$ and $\beta$ glycosidic linkages. $\alpha$ linkages are found in starch (digestible); $\beta$ linkages are found in cellulose (not digestible by humans).
Thinking all lipids are fats. Steroids and phospholipids are also lipids.
Confusing saturated and unsaturated fats. Saturated fats have no double bonds and are solid; unsaturated fats have double bonds and are liquid.
Misidentifying enzyme inhibition effects. Competitive inhibition increases $K_m$ but does not change $V_{\max}$ ; noncompetitive decreases $V_{\max}$ but does not change $K_m$ .
Confusing DNA and RNA. DNA has deoxyribose and thymine; RNA has ribose and uracil.
Thinking enzymes change $\Delta G$ . Enzymes only lower activation energy; they do not change the thermodynamics of the reaction.
Confusing purines and pyrimidines. Purines (A, G) have two rings; pyrimidines (C, T, U) have one ring. “Pure As Gold” for purines.
Forgetting that denaturation only affects secondary, tertiary, and quaternary structure. The primary structure (amino acid sequence) is not changed by denaturation.
Confusing exergonic and endergonic. Exergonic: $\Delta G \lt 0$ Releases energy. Endergonic: $\Delta G \gt 0$ Requires energy.

Practice Questions

Compare and contrast starch, glycogen, and cellulose in terms of structure and function.
Explain how the properties of water (cohesion, adhesion, high specific heat) contribute to the ability of plants to transport water from roots to leaves.
Draw the general structure of an amino acid and identify the components that differ among the 20 common amino acids.
Describe the four levels of protein structure and the types of bonds that stabilize each level.
Explain why a competitive inhibitor increases $K_m$ but does not change $V_{\max}$ While a noncompetitive inhibitor decreases $V_{\max}$ but does not change $K_m$ .
Compare DNA and RNA in terms of structure, bases, sugar, and function.
Explain how ATP hydrolysis can drive an endergonic reaction.
A researcher observes that an enzyme has maximum activity at pH 7.0 and $37^\circ\mathrm{C$ . Explain what happens to the reaction rate if the pH is changed to 10.0 and why.
Calculate the pH of a solution with $[\mathrm{H^+] = 2.5 \times 10^{-3}$ M.
A DNA molecule is found to be 24% cytosine. Using Chargaff’s rules, determine the percentage of all four bases.
Explain why ice is less dense than liquid water and describe the ecological significance of this property.
Describe the induced-fit model of enzyme action and explain how it differs from the lock-and-key model.
A triglyceride is hydrolysed. Name the products and describe the chemical bonds that are broken.
Explain how the buffer system in blood maintains pH close to 7.4 during exercise.
Calculate the $\Delta G$ of a coupled reaction where the endergonic step has $\Delta G = +17$ kJ/mol and the exergonic step (ATP hydrolysis) has $\Delta G = -30.5$ kJ/mol. Is the overall reaction spontaneous?
Explain why phospholipids spontaneously form bilayers in water.
Describe the role of chaperone proteins in protein folding and explain why their function is important.
Compare the structure and function of DNA polymerase and RNA polymerase.
Explain why enzymes are sensitive to pH changes, with reference to the effect on amino acid side chains.
A student measures enzyme activity at different temperatures and finds that the rate peaks at $40^\circ\mathrm{C$ and drops to zero at $60^\circ\mathrm{C$ . Explain these results with reference to protein structure.

Review: Macromolecule Identification Tests

Several biochemical tests are used to identify the presence of specific macromolecules in samples. These are commonly tested on the AP exam.

Test	Reagent	Positive Result	Macromolecule Detected
Benedict’s test	Benedict’s solution	Brick-red precipitate (when heated)	Reducing sugars
Iodine test	Iodine solution	Blue-black colour	Starch
Biuret test	Biuret reagent	Violet/purple colour	Protein
Emulsion test	Ethanol	Cloudy white emulsion layer	Lipids
Sudan III test	Sudan III stain	Red-stained oil layer	Lipids

Worked Example: Designing an identification experiment.

A student has three unknown solutions (A, B, and C) and needs to identify which contains starch, Which contains protein, and which contains glucose.

Add iodine solution to each: the one that turns blue-black contains starch.
Add Benedict’s solution to each of the remaining two and heat: the one that produces a brick-red precipitate contains glucose.
Add Biuret reagent to the last remaining solution: a violet colour indicates protein.

Worked Example: Interpreting Benedict’s test results.

A student performs Benedict’s test on two solutions. Solution X produces a brick-red precipitate. Solution Y produces a blue colour (no change).

Interpretation: Solution X contains a reducing sugar (e.g., glucose, maltose, or fructose). The Brick-red precipitate is copper(I) oxide, formed when copper(II) ions in Benedict’s solution are Reduced by the sugar. Solution Y does not contain reducing sugars (it may contain a non-reducing Sugar such as sucrose, or no sugar at all). To test for non-reducing sugars, the sample would first Be hydrolysed with dilute acid, neutralised, and then tested with Benedict’s solution.

Review: Levels of Structural Organisation in Proteins

Understanding the four levels of protein structure is essential for the AP exam, particularly how Disruption of each level affects function.

Primary structure: The specific sequence of amino acids, determined by the nucleotide sequence Of the gene. Even a single amino acid substitution (point mutation) can alter protein function. Example: sickle cell disease results from a single substitution of glutamic acid with valine at Position 6 of the beta-globin chain.

Secondary structure: Local folding into $\alpha$ -helices and $\beta$ -pleated sheets, stabilised By hydrogen bonds between the backbone amino and carboxyl groups (not the R groups). These Structures are common in fibrous proteins (e.g., keratin in hair contains many $\alpha$ -helices).

Tertiary structure: The overall three-dimensional shape of a single polypeptide, stabilised by Interactions between R groups: hydrogen bonds, ionic bonds (salt bridges), hydrophobic interactions (nonpolar R groups cluster in the interior away from water), and disulfide bridges (covalent bonds Between cysteine residues). The tertiary structure determines the protein’s function.

Quaternary structure: The assembly of two or more polypeptide subunits into a functional Protein. Example: haemoglobin has four subunits (two alpha and two beta chains). Each subunit has Its own tertiary structure, and the subunits interact to form the quaternary structure.

Worked Example: Effect of pH on protein structure.

Pepsin (a digestive enzyme in the stomach) has an optimum pH of approximately 2. If pepsin is placed In a neutral solution (pH 7), its activity decreases dramatically. This is because the change in pH Alters the ionisation of amino acid side chains (R groups). Ionic bonds and hydrogen bonds that Depend on specific charges are disrupted, causing the protein to unfold (denature). The active site Loses its specific shape, and the enzyme can no longer bind its substrate.

Review: Properties of Carbon

Carbon is the backbone of organic molecules because it can form four covalent bonds (tetrahedral Geometry), allowing it to create complex, diverse structures. Key concepts:

Carbon can form single, double, and triple bonds.
Carbon chains can be straight, branched, or ring-shaped.
Carbon can bond with many other elements (H, O, N, P, S).
Isomers are molecules with the same molecular formula but different structures:
Structural isomers: Different arrangement of the carbon skeleton (e.g., butane vs isobutane).
Geometric isomers (cis-trans): Different spatial arrangement around a double bond.
Enantiomers: Mirror images of each other (chirality). Enantiomers can have very different biological activities (e.g., one enantiomer of thalidomide is therapeutic, the other is teratogenic).

Worked Example: Why carbon’s bonding versatility matters.

The ability of carbon to form four bonds allows the construction of an enormous variety of organic Molecules. For example, a chain of just 20 carbon atoms can be arranged in over 300,000 different Structural isomers. This diversity is the foundation of the complexity of biological molecules (proteins, nucleic acids, carbohydrates, and lipids), each with specific structures that determine Their functions.

Review: Nucleic Acid Structure and Replication

DNA replication (semiconservative):

Initiation: Helicase unwinds the double helix at the origin of replication. Single-strand binding proteins prevent the strands from re-annealing. Topoisomerase relieves tension ahead of the replication fork.
Elongation: DNA polymerase III adds nucleotides to the 3’ end of the growing strand (DNA synthesis is always 5’ to 3’). The leading strand is synthesised continuously; the lagging strand is synthesised in short Okazaki fragments, which are later joined by DNA ligase. RNA primase synthesises a short RNA primer to initiate each fragment.
Termination: The two daughter DNA molecules each contain one original strand and one new strand (semiconservative replication).

Evidence for semiconservative replication: The Meselson-Stahl experiment (1958) used heavy Nitrogen ( $^{15}\mathrm{N$ ) to label DNA. After one round of replication in $^{14}\mathrm{N$ Medium, all DNA molecules contained one heavy strand and one light strand (intermediate density), Ruling out conservative replication. After two rounds, both light and intermediate density DNA was Observed, confirming the semiconservative model.

Review: Lipids and Cell Membrane Structure in Detail

Phospholipid structure: Each phospholipid molecule has a hydrophilic (water-loving) phosphate Head and two hydrophobic (water-fearing) fatty acid tails. In aqueous solution, phospholipids Spontaneously form bilayers, with the hydrophilic heads facing outward towards water and the Hydrophobic tails facing inward, away from water. This self-assembly is driven by the hydrophobic Effect.

Saturated vs unsaturated fatty acids:

Saturated fatty acids: No carbon-carbon double bonds. The hydrocarbon chains are straight and can pack tightly together. This makes the membrane more rigid. Found primarily in animal fats.
Unsaturated fatty acids: One or more carbon-carbon double bonds, creating kinks in the chain. These kinks prevent tight packing, increasing membrane fluidity. Found primarily in plant oils.
Trans fats: Artificially hydrogenated unsaturated fats. The hydrogenation process straightens the kinks, making them behave like saturated fats. Associated with increased risk of cardiovascular disease because they raise LDL cholesterol and lower HDL cholesterol.

Cholesterol in the membrane: Cholesterol is a steroid molecule embedded in the hydrophobic Interior of the phospholipid bilayer. It acts as a fluidity buffer:

At high temperatures: cholesterol restrains the movement of phospholipids, reducing fluidity and preventing the membrane from becoming too permeable.
At low temperatures: cholesterol prevents tight packing of phospholipids, maintaining fluidity and preventing the membrane from solidifying.

Worked Example: Why phospholipids form bilayers but not micelles in cell membranes.

Phospholipids with two fatty acid tails (as found in cell membranes) are cylindrical in shape. These Cylindrical molecules pack together most efficiently in a bilayer arrangement. Phospholipids with One fatty acid tail (lysophospholipids) are cone-shaped and form micelles (small spherical Structures) instead. The bilayer arrangement creates a stable, flexible membrane with two distinct Layers, which is essential for the function of cell membranes.

Review: Carbohydrate Chemistry in Detail

Monosaccharides: The simplest carbohydrates, with the general formula $\mathrm{C_n(\mathrm{H_2\mathrm{O)_n$ . Glucose ( $\mathrm{C_6\mathrm{H_{12}\mathrm{O_6$ ) exists In two isomeric forms: alpha ( $\alpha$ ) and beta ( $\beta$ ) glucose. The difference is the position Of the hydroxyl (-OH) group on carbon 1. In $\alpha$ -glucose, the -OH is below the plane of the Ring; in $\beta$ -glucose, it is above. This apparently small difference has enormous consequences: $\alpha$ -glucose forms starch (digestible) and glycogen, while $\beta$ -glucose forms cellulose (indigestible by most animals).

Disaccharides: Formed by condensation reactions between two monosaccharides:

Disaccharide	Component monosaccharides	Bond type	Found in
Maltose	Glucose + glucose	$\alpha$ -1,4	Germinating seeds
Sucrose	Glucose + fructose	$\alpha$ -1,2	Sugar cane, beet
Lactose	Galactose + glucose	$\beta$ -1,4	Milk

Why lactose intolerance occurs: Lactase, the enzyme that hydrolyses lactose into glucose and Galactose, is produced in the small intestine of infants. In many populations, lactase production Decreases after weaning. Without lactase, lactose passes undigested into the large intestine, where Bacteria ferment it, producing gas and causing bloating, cramps, and diarrhoea. This condition is Called lactose intolerance and is particularly common in East Asian, African, and Native American Populations.

Worked Example: Benedict’s test for reducing and non-reducing sugars.

A student tests two solutions with Benedict’s reagent. Solution A produces a brick-red precipitate When heated. Solution B shows no change. The student then boils Solution B with dilute hydrochloric Acid, neutralises it, and tests again with Benedict’s reagent. This time, Solution B produces a Brick-red precipitate.

Explanation: Solution A contains a reducing sugar (e.g., glucose, maltose) that can directly Reduce copper(II) ions in Benedict’s reagent to copper(I) oxide. Solution B contains a non-reducing Sugar (e.g., sucrose). Sucrose must first be hydrolysed into its component monosaccharides (glucose And fructose) by acid hydrolysis. After hydrolysis, the resulting monosaccharides are reducing Sugars and give a positive Benedict’s test.

Review: Amino Acid Chemistry

Essential vs non-essential amino acids: Humans can synthesise 11 of the 20 amino acids (non-essential amino acids). The remaining 9 must be obtained from the diet (essential amino acids: Histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, valine).

Zwitterions: At physiological pH (approximately 7.4), amino acids exist as zwitterions — Molecules with both a positive charge (on the amino group, $-\mathrm{NH_3^+$ ) and a negative charge (on the carboxyl group, $-\mathrm{COO^-$ ). The overall charge is neutral.

Peptide bond formation: A condensation reaction between the carboxyl group of one amino acid and The amino group of another forms a peptide bond (covalent bond) and releases a water molecule. A Dipeptide is formed from two amino acids; a polypeptide is a chain of many amino acids.

Protein diversity: With 20 different amino acids, the number of possible polypeptides of length $n$ is $20^n$ . For a protein of 100 amino acids, there are $20^{100}$ possible sequences (an Astronomically large number). This explains how a relatively small set of building blocks can Generate the enormous diversity of proteins found in living organisms.

Review: Enzyme Regulation and Allosteric Regulation

Allosteric regulation: Many enzymes have allosteric sites (sites distinct from the active site) Where regulatory molecules can bind. Binding of an allosteric activator stabilises the active Conformation of the enzyme, increasing its activity. Binding of an allosteric inhibitor stabilises The inactive conformation, decreasing its activity.

Cooperativity: In some multi-subunit enzymes, binding of a substrate to one active site Increases the affinity of the other active sites for the substrate. This produces a sigmoidal (S-shaped) velocity-vs-substrate curve rather than the hyperbolic curve predicted by the Michaelis-Menten equation. Haemoglobin (which is not an enzyme but a transport protein) shows Cooperative binding of oxygen: binding of the first $\mathrm{O_2$ molecule increases the affinity For subsequent $\mathrm{O_2$ molecules.

Feedback inhibition: A common form of metabolic regulation where the end product of a pathway Inhibits an enzyme early in the pathway. This prevents wasteful overproduction of the end product. For example, in the pathway for isoleucine synthesis, isoleucine acts as an allosteric inhibitor of Threonine deaminase, the first enzyme in the pathway.

Review: Experimental Design for Enzyme Investigations

Variables to control:

Temperature: Use a water bath to maintain a constant temperature. Allow time for the enzyme and substrate to reach the bath temperature before starting the reaction.
pH: Use a buffer solution to maintain a constant pH.
Substrate concentration: Prepare solutions of known concentration using a volumetric flask.
Enzyme concentration: Use the same volume and concentration of enzyme solution in each trial.
Volume: Use the same total volume of reaction mixture in each trial.
Timing: Start the clock at the moment of mixing and stop it at a consistent end point (e.g., colour change, gas production).

Reliability: Repeat each measurement at least three times and calculate a mean. Identify and Exclude anomalous results (those that are inconsistent with the other repeats).

Validity: Ensure that the experiment actually measures what it claims to measure. Control all Relevant variables to ensure that only the independent variable affects the dependent variable.

Review: Water’s Role in Temperature Regulation

Water’s high specific heat capacity ( $4.18$ J/g/ $\degree$ C) means it can absorb or release large Amounts of heat with relatively small changes in temperature. This property has important biological Consequences:

Oceans moderate coastal climates: Because water heats and cools slowly, coastal areas have smaller temperature fluctuations than inland areas at the same latitude.
Organisms maintain stable internal temperatures: The high water content of organisms ( 60—90%) contributes to thermal stability.
Evaporative cooling: Water’s high heat of vaporisation ( $2260$ J/g at $100\degree$ C) means that sweating (in mammals) and transpiration (in plants) are very effective cooling mechanisms.

Worked Example: A person running on a hot day produces approximately 500 mL of sweat per hour. To evaporate 500 mL of water requires approximately $500 \times 2260 = 1,130,000$ J (1130 kJ) of Energy. This energy is drawn from the body as heat, significantly reducing body temperature and Preventing overheating.

Review: The Importance of Functional Groups in Biological Molecules

Functional groups are specific groups of atoms within molecules that determine the characteristics Of the molecule and how it reacts.

Functional Group	Structure	Found in	Properties
Hydroxyl (-OH)	-OH	Alcohols, sugars, amino acids	Polar; forms hydrogen bonds
Carbonyl (C=O)	C=O	Aldehydes, ketones, sugars	Polar; reactive
Carboxyl (-COOH)	-COOH	Amino acids, fatty acids	Acts as an acid; releases $\mathrm{H^+$
Amino (-NH $_2$ )	-NH $_2$	Amino acids	Acts as a base; accepts $\mathrm{H^+$
Phosphate	-PO $_4$	ATP, DNA, phospholipids	Negative charge; energy transfer
Sulfhydryl (-SH)	-SH	Cysteine (amino acid)	Forms disulfide bridges in proteins

Worked Example: The role of functional groups in amino acid behaviour.

Amino acids contain both an amino group (-NH $_2$ Basic) and a carboxyl group (-COOH, acidic). In Aqueous solution at physiological pH, the amino group accepts a proton (becoming -NH $_3^+$ ) and the Carboxyl group donates a proton (becoming -COO $^-$ ), forming a zwitterion with both a positive and a Negative charge. The R group (side chain) determines the specific properties of each amino acid: Nonpolar R groups are hydrophobic, while polar and charged R groups are hydrophilic. Cysteine Contains a sulfhydryl group (-SH) that can form disulfide bridges with another cysteine, creating Covalent cross-links that stabilise protein structure.

Practice Problems

Question 1: Chargaff's rules and DNA composition

A double-stranded DNA molecule is found to be $18\%$ adenine. Calculate the percentage of each of The other three bases. If the DNA contains $10,000$ base pairs, how many hydrogen bonds hold the two Strands together?

Answer

By Chargaff’s rules, $[\mathrm{A] = [\mathrm{T] = 18\%$ .

Since $\mathrm{A + \mathrm{T + \mathrm{G + \mathrm{C = 100\%$ : $\mathrm{G + \mathrm{C = 100 - 18 - 18 = 64\%$ .

By Chargaff’s rules, $[\mathrm{G] = [\mathrm{C] = 64 / 2 = 32\%$ .

So: $\mathrm{A = 18\%$$\mathrm{T = 18\%$$\mathrm{G = 32\%$$\mathrm{C = 32\%$ .

Hydrogen bonds: A-T pairs have 2 H-bonds, G-C pairs have 3 H-bonds.

Number of A-T pairs: $0.18 \times 10,000 = 1,800$ pairs, contributing $1,800 \times 2 = 3,600$ H-bonds.

Number of G-C pairs: $0.32 \times 10,000 = 3,200$ pairs, contributing $3,200 \times 3 = 9,600$ H-bonds.

Total hydrogen bonds: $3,600 + 9,600 = 13,200$ .

Question 2: Enzyme kinetics with an inhibitor

An enzyme has $K_m = 4 \mathrm{ \mu M$ and $V_{\max} = 20 \mathrm{ \mu mol/min$ . In the presence Of a competitive inhibitor at $8 \mathrm{ \mu M$ The apparent $K_m$ increases to $12 \mathrm{ \mu M$ . Calculate the reaction velocity at a substrate concentration of $6 \mathrm{ \mu M$ both with and without the inhibitor.

Answer

Without inhibitor: $v = \frac{V_{\max}[S]}{K_m + [S]} = \frac{20 \times 6}{4 + 6} = \frac{120}{10} = 12 \mathrm{ \mu mol/min$ .

With inhibitor: $v = \frac{V_{\max}[S]}{K_m^{app} + [S]} = \frac{20 \times 6}{12 + 6} = \frac{120}{18} = 6.67 \mathrm{ \mu mol/min$ .

The competitive inhibitor reduces the reaction velocity from $12 \mathrm{ \mu mol/min$ to $6.67 \mathrm{ \mu mol/min$ at this substrate concentration. Note that $V_{\max}$ is unchanged; at Very high substrate concentrations, both velocities would approach $20 \mathrm{ \mu mol/min$ .

Question 3: pH and buffer calculations

A buffer solution contains $0.10 \mathrm{ M$ acetic acid ( $\mathrm{CH_3\mathrm{COOH$ $\mathrm{pK_a = 4.76$ ) and $0.20 \mathrm{ M$ sodium acetate ( $\mathrm{CH_3\mathrm{COO^-$ ). Calculate the pH of this buffer. If $0.01 \mathrm{ mol$ of $\mathrm{HCl$ is added to $1.0 \mathrm{ L$ of this buffer, what is the new pH?

Answer

Using the Henderson-Hasselbalch equation:

$\mathrm{pH = \mathrm{pK_a + \log\frac{[\mathrm{A^-]}{[\mathrm{HA]} = 4.76 + \log\frac{0.20}{0.10} = 4.76 + \log(2) = 4.76 + 0.301 = 5.06$ .

After adding $0.01 \mathrm{ mol$ of $\mathrm{HCl$ to $1.0 \mathrm{ L$ :

The $\mathrm{HCl$ reacts with the conjugate base: $\mathrm{CH_3\mathrm{COO^- + \mathrm{H^+ \to \mathrm{CH_3\mathrm{COOH$ .

New $[\mathrm{CH_3\mathrm{COO^-] = 0.20 - 0.01 = 0.19 \mathrm{ M$ .

New $[\mathrm{CH_3\mathrm{COOH] = 0.10 + 0.01 = 0.11 \mathrm{ M$ .

New $\mathrm{pH = 4.76 + \log\frac{0.19}{0.11} = 4.76 + \log(1.727) = 4.76 + 0.237 = 4.997 \approx 5.00$ .

The pH changed by only $0.06$ units, demonstrating the buffer’s effectiveness.

Question 4: Energy coupling and thermodynamics

The phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate has $\Delta G = +16.7 \mathrm{ kJ/mol$ under cellular conditions. ATP hydrolysis has $\Delta G = -30.5 \mathrm{ kJ/mol$ . Calculate the overall $\Delta G$ when these reactions are Coupled. Is the overall reaction spontaneous?

Answer

Overall $\Delta G = +16.7 + (-30.5) = -13.8 \mathrm{ kJ/mol$ .

Since $\Delta G \lt 0$ The overall coupled reaction is spontaneous. The energy released by ATP Hydrolysis ( $-30.5 \mathrm{ kJ/mol$ ) more than compensates for the energy required for the Phosphorylation ( $+16.7 \mathrm{ kJ/mol$ ). This is how phosphofructokinase drives the committed Step of glycolysis forward. The excess energy ( $13.8 \mathrm{ kJ/mol$ ) is released as heat.

Question 5: Protein structure and denaturation

A researcher discovers that a protein loses its biological activity when heated to $45^\circ\mathrm{C$ But regains full activity when cooled back to $25^\circ\mathrm{C$ . When Heated to $70^\circ\mathrm{C$ The protein permanently loses activity even after cooling. Explain These observations with reference to the levels of protein structure.

Answer

At $45^\circ\mathrm{C$ The protein undergoes reversible denaturation. The secondary and tertiary Structures are disrupted (hydrogen bonds, hydrophobic interactions, and ionic bonds break), causing The protein to unfold and lose its active site shape. However, the primary structure (amino acid Sequence) remains intact. When cooled, these non-covalent interactions can re-form, allowing the Protein to refold into its native, functional conformation.

At $70^\circ\mathrm{C$ The denaturation is irreversible. At this higher temperature, the protein May aggregate (unfolded proteins expose hydrophobic regions that stick together), or disulfide Bridges (covalent bonds) may be disrupted. Once aggregated or covalently damaged, the protein cannot Refold correctly upon cooling because the protein is trapped in incorrect, aggregated conformations. The loss of disulfide bridges is particularly significant because these covalent cross-links cannot Spontaneously re-form in the correct positions without chaperone proteins.

Summary

This topic covers the essential chemistry of chemistry of life, including key reactions, underlying theories, and practical applications.

Key concepts include:

ionic, covalent, and metallic bonding
electronegativity and bond polarity
intermolecular forces
giant and simple molecular structures
VSEPR theory

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

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