Biological Molecules

Biological Molecules

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Board Coverage AQA Paper 1 | Edexcel A Paper 1 | OCR (A) Paper 1 | CIE Paper 2

1. Water

1.1 Structure and Properties

Water ($\mathrm{H_2O}$) is a polar molecule. The oxygen atom is more electronegative than hydrogen, creating a dipole with $\delta^-$ on oxygen and $\delta^+$ on each hydrogen. This polarity underpins water's biological significance.

Key properties of water:

Property	Cause	Biological Significance
High specific heat capacity ($4.18\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$)	Hydrogen bonding absorbs energy before temperature rises	Temperature buffers for organisms; aquatic environments are thermally stable
High latent heat of evaporation ($2.26\ \mathrm{kJ\ g^{-1}}$)	Hydrogen bonds must break for evaporation	Effective cooling through sweating
Cohesion and surface tension	Hydrogen bonding between water molecules	Water transport in xylem columns; surface habitat for some organisms
High solvent power	Polarity allows interaction with ions and polar molecules	Medium for metabolic reactions; transport medium in blood and phloem
Lower density of ice	Hydrogen bonding creates an open lattice in solid state	Insulating layer on water bodies; aquatic organisms survive winter
Transparency	Light passes through water	Photosynthesis can occur below the surface

1.2 Hydrogen Bonding in Water

Each water molecule can form up to four hydrogen bonds: two through its hydrogen atoms (donor) and two through lone pairs on oxygen (acceptor). This extensive network gives water its anomalously high boiling point ($100\ ^\circ\mathrm{C}$) compared to similarly sized molecules such as $\mathrm{H_2S}$ ($-60\ ^\circ\mathrm{C}$).

warning

Common Pitfall Students often confuse hydrogen bonds with covalent bonds. Hydrogen bonds are intermolecular forces -- they are much weaker ($\approx 20\ \mathrm{kJ\ mol^{-1}}$) than covalent bonds ($\approx 350\ \mathrm{kJ\ mol^{-1}}$). They form between a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) and a lone pair on a neighbouring electronegative atom.

2. Carbohydrates

2.1 Monosaccharides

Monosaccharides are the monomers of carbohydrates with the general formula $(\mathrm{CH_2O})_n$. The most biologically important monosaccharide is glucose, $\mathrm{C_6H_{12}O_6}$.

Glucose exists as two isomers that differ at carbon 1:

• $\alpha$-glucose: the $-\mathrm{OH}$ group on $\mathrm{C_1}$ is below the plane of the ring.
• $\beta$-glucose: the $-\mathrm{OH}$ group on $\mathrm{C_1}$ is above the plane of the ring.

This seemingly minor difference has profound structural consequences: $\alpha$-glucose forms starch and glycogen, while $\beta$-glucose forms cellulose -- two polymers with entirely different properties.

2.2 Disaccharides

Disaccharides are formed by a condensation reaction (glycosidic bond formation) between two monosaccharides, releasing one molecule of water:

$\mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6} \longrightarrow \mathrm{C_{12}H_{22}O_{11}} + \mathrm{H_2O}$

Disaccharide	Component Monosaccharides	Glycosidic Bond
Maltose	$\alpha$-glucose + $\alpha$-glucose	$\alpha$-1,4
Sucrose	$\alpha$-glucose + fructose	$\alpha$-1,2
Lactose	$\alpha$-glucose + galactose	$\beta$-1,4

Hydrolysis of disaccharides requires addition of water and is catalysed by specific enzymes (maltase, sucrase, lactase respectively).

2.3 Polysaccharides

Polysaccharides are long polymers of monosaccharides joined by glycosidic bonds. Three are particularly important:

Starch is the primary energy storage molecule in plants. It consists of two components:

• Amylose: unbranched chain of $\alpha$-glucose with $\alpha$-1,4 glycosidic bonds. The chain coils into a helix, making it compact and insoluble -- ideal for storage.
• Amylopectin: branched chain of $\alpha$-glucose with $\alpha$-1,4 bonds and $\alpha$-1,6 branch points every 24--30 glucose units. The branches create many free ends for enzyme action.

Glycogen is the animal equivalent of starch, found in liver and muscle cells. It is more densely branched (every 8--12 glucose units) than amylopectin, allowing faster hydrolysis to meet the higher metabolic demands of animals.

Cellulose is a structural polysaccharide found in plant cell walls. It consists of long, unbranched chains of $\beta$-glucose joined by $\beta$-1,4 glycosidic bonds. Every other glucose molecule is rotated $180^\circ$, allowing adjacent chains to form extensive hydrogen bonds. These chains bundle into microfibrils, which provide tremendous tensile strength.

warning

warning has $\beta$-1,4 glycosidic bonds. The $\beta$ configuration is what causes the alternate rotation and enables the hydrogen bonding between chains that gives cellulose its strength.

2.4 Testing for Carbohydrates

Test	Reducing sugars	Non-reducing sugars	Starch
Reagent	Benedict's reagent	Benedict's reagent (after acid hydrolysis)	Iodine in potassium iodide
Method	Add reagent, heat in water bath at $80\ ^\circ\mathrm{C}$ for 5 min	Hydrolyse with dilute $\mathrm{HCl}$, neutralise with $\mathrm{NaHCO_3}$, then Benedict's	Add iodine solution directly
Positive result	Brick-red precipitate	Brick-red precipitate (after hydrolysis)	Blue-black colour

The colour of the Benedict's test indicates sugar concentration: blue (trace) $\to$ green $\to$ yellow $\to$ orange $\to$ brick red (high concentration).

3. Lipids

3.1 Triglycerides

Triglycerides are formed by a condensation reaction between one molecule of glycerol (a 3-carbon alcohol with three $-\mathrm{OH}$ groups) and three molecules of fatty acids (long hydrocarbon chains with a terminal $-\mathrm{COOH}$ group). Each ester bond releases one molecule of water:

$\mathrm{glycerol} + 3\ \mathrm{fatty\ acids} \longrightarrow \mathrm{triglyceride} + 3\ \mathrm{H_2O}$

Fatty acids are classified by the presence of double bonds in their hydrocarbon chain:

• Saturated fatty acids: no $\mathrm{C=C}$ double bonds; straight chains pack tightly; solid at room temperature (e.g., in animal fats).
• Unsaturated fatty acids: one or more $\mathrm{C=C}$ double bonds; kinked chains cannot pack tightly; liquid at room temperature (e.g., in plant oils). Monounsaturated = one double bond; polyunsaturated = two or more.

3.2 Phospholipids

Phospholipids are similar to triglycerides but with one fatty acid replaced by a phosphate group attached to a small organic molecule (e.g., choline). This gives phospholipids a hydrophilic head (phosphate group) and two hydrophobic tails (fatty acid chains). This amphipathic nature is fundamental to the formation of cell membranes.

In aqueous solution, phospholipids spontaneously form bilayers: the hydrophilic heads face outward towards water, while the hydrophobic tails face inward, shielded from water. This is a direct consequence of the hydrophobic effect and the second law of thermodynamics.

3.3 Roles of Lipids

• Energy storage: triglycerides contain more than twice the energy per unit mass than carbohydrates ($\approx 39\ \mathrm{kJ\ g^{-1}}$ vs. $\approx 17\ \mathrm{kJ\ g^{-1}}$).
• Insulation: subcutaneous fat reduces heat loss.
• Protection: adipose tissue cushions organs.
• Membrane structure: phospholipid bilayers form the basis of all cell membranes.
• Hormone precursors: cholesterol is a precursor for steroid hormones (testosterone, oestrogen).

3.4 Testing for Lipids

The emulsion test: dissolve the sample in ethanol, then pour into water. Lipids are insoluble in water but soluble in ethanol. A cloudy white emulsion indicates the presence of lipids.

4. Proteins

4.1 Amino Acids

Proteins are polymers of amino acids. There are 20 standard amino acids, all sharing a common structure:

$\mathrm{H_2N{-}CHR{-}COOH}$

where $R$ is the variable side chain (r-group) that determines the amino acid's chemical properties.

Amino acids are amphoteric: they can act as both acids and bases because they contain both $-\mathrm{NH_2}$ (basic) and $-\mathrm{COOH}$ (acidic) groups. At physiological pH ($\approx 7.4$), amino acids exist as zwitterions with $-\mathrm{NH_3^+}$ and $-\mathrm{COO^-}$.

4.2 Peptide Bonds

Amino acids are joined by condensation reactions forming peptide bonds:

$$\mathrm{H_2N{-}CHR_1{-}COOH} + \mathrm{H_2N{-}CHR_2{-}COOH} \longrightarrow \mathrm{H_2N{-}CHR_1{-}CO{-}NH{-}CHR_2{-}COOH} + \mathrm{H_2O}$$

A dipeptide has two amino acids; a polypeptide has many. A protein is one or more polypeptides folded into a specific 3D conformation.

4.3 Levels of Protein Structure

Primary structure: the sequence of amino acids in the polypeptide chain, joined by peptide bonds. This is determined directly by the gene sequence.

Secondary structure: local folding patterns stabilised by hydrogen bonds between backbone $-\mathrm{C=O}$ and $-\mathrm{N-H}$ groups (not side chains). Two main types:

• $\alpha$-helix: a right-handed coil with hydrogen bonds parallel to the axis.
• $\beta$-pleated sheet: hydrogen bonds between adjacent extended strands.

Tertiary structure: the overall 3D shape of a single polypeptide chain, stabilised by:

• Hydrogen bonds
• Ionic (electrostatic) bonds between charged $R$ groups
• Disulfide bridges (covalent $-\mathrm{S-S-}$ bonds between cysteine residues)
• Hydrophobic interactions (non-polar $R$ groups cluster in the interior)
• Van der Waals forces

Quaternary structure: the arrangement of multiple polypeptide chains (subunits) into a functional protein. Examples: haemoglobin (4 subunits), immunoglobulins (4 subunits).

4.4 Protein Types

• Globular proteins: compact, spherical, soluble in water. Enzymes, antibodies, haemoglobin, hormones (insulin).
• Fibrous proteins: elongated, structural, insoluble. Collagen (connective tissue), keratin (hair, nails), elastin (artery walls).

4.5 The Biuret Test

Add Biuret reagent (copper(II) sulfate in alkaline solution) to the sample. A colour change from blue to violet/purple indicates the presence of peptide bonds, hence protein. The intensity of the colour is proportional to protein concentration.

warning

warning bonds. A solution of free amino acids will not give a positive Biuret test (or only a very weak one for dipeptides).

5. Nucleic Acids

5.1 DNA Structure

Deoxyribonucleic acid (DNA) is a double-stranded polymer of nucleotides. Each nucleotide consists of:

1. A deoxyribose sugar (a pentose with $-\mathrm{H}$ at the $2'$ position)
2. A phosphate group
3. A nitrogenous base: adenine (A), thymine (T), cytosine (C), or guanine (G)

Nucleotides are linked by phosphodiester bonds between the $3'$ carbon of one sugar and the $5'$ carbon of the next, forming a sugar-phosphate backbone with a directional $5' \to 3'$ polarity.

Chargaff's rules: in double-stranded DNA, $[\mathrm{A}] = [\mathrm{T}]$ and $[\mathrm{C}] = [\mathrm{G}]$. This is a consequence of complementary base pairing: A pairs with T (2 hydrogen bonds), C pairs with G (3 hydrogen bonds).

Watson-Crick model: the two strands run antiparallel ($5' \to 3'$ alongside $3' \to 5'$) and twist into a right-handed double helix with approximately 10 base pairs per turn.

5.2 RNA Structure

Ribonucleic acid (RNA) differs from DNA in three respects:

1. The sugar is ribose (with $-\mathrm{OH}$ at the $2'$ position)
2. The base thymine is replaced by uracil (U)
3. RNA is typically single-stranded (though it can fold back on itself to form secondary structures)

Three types of RNA:

Type	Function	Structure
Messenger RNA (mRNA)	Carries genetic code from DNA to ribosomes	Single-stranded, short-lived
Transfer RNA (tRNA)	Carries amino acids to ribosomes	Cloverleaf shape; anticodon loop
Ribosomal RNA (rRNA)	Structural and catalytic component of ribosomes	Large, complex, folded

5.3 ATP

Adenosine triphosphate ($\mathrm{ATP}$) is the universal energy currency of the cell:

$\mathrm{ATP} + \mathrm{H_2O} \longrightarrow \mathrm{ADP} + P_i + \mathrm{energy}$

$\Delta G \approx -30.5\ \mathrm{kJ\ mol^{-1}}$

The hydrolysis of the terminal phosphoanhydride bond is exergonic and releases energy that drives endergonic cellular processes. ATP is not a long-term energy store; it is a short-term, immediate energy transfer molecule that must be continuously regenerated through cellular respiration or photosynthesis.

warning

warning starch (plants) are the long-term energy stores. ATP is regenerated within seconds of use.

6. Enzymes

6.1 Enzyme Structure and Function

Enzymes are biological catalysts -- globular proteins that increase the rate of metabolic reactions by lowering the activation energy ($\Delta G^\ddagger$) without being consumed.

The active site is a specific 3D region of the enzyme where the substrate binds. It is complementary to the substrate in shape, charge, and solubility -- analogous to a lock and key.

6.2 The Induced-Fit Model

The induced-fit model (Koshland, 1958) refined the original lock-and-key hypothesis. Upon substrate binding, the active site changes conformation to mould more tightly around the substrate. This:

1. Places strain on substrate bonds, making them easier to break
2. Positions catalytic residues optimally for the reaction
3. Excludes water molecules that could interfere with the reaction

6.3 Enzyme Kinetics

The rate of an enzyme-catalysed reaction ($v$) depends on substrate concentration $[S]$:

$v = \frac◆LB◆V_{\max}[S]◆RB◆◆LB◆K_m + [S]◆RB◆$

where:

• $V_{\max}$ is the maximum rate when all active sites are saturated
• $K_m$ (the Michaelis constant) is the substrate concentration at which $v = V_{\max}/2$

A low $K_m$ indicates high affinity between enzyme and substrate (the enzyme reaches half $V_{\max}$ at low $[S]$). A high $K_m$ indicates low affinity.

6.4 Factors Affecting Enzyme Activity

Temperature: as temperature increases, kinetic energy increases, raising the rate of collisions between enzyme and substrate. Above the optimum temperature (approximately $37\ ^\circ\mathrm{C}$ for human enzymes), the protein denatures -- hydrogen bonds and other weak interactions break, the active site changes shape, and the enzyme loses function. Denaturation is irreversible.

pH: each enzyme has an optimum pH at which its active site has the correct conformation. Extreme pH alters the ionisation of $R$ groups, disrupting ionic bonds and hydrogen bonds that maintain tertiary structure. Pepsin has an optimum pH of $\approx 2$ (stomach); salivary amylase has an optimum pH of $\approx 7$ (mouth).

Substrate concentration: at low $[S]$, rate is proportional to $[S]$ (first-order kinetics). At high $[S]$, all active sites are occupied and the rate plateaus at $V_{\max}$ (zero-order kinetics).

Enzyme concentration: at fixed $[S]$, rate is proportional to $[E]$ provided substrate is not limiting.

6.5 Enzyme Inhibition

Competitive inhibition: the inhibitor has a similar structure to the substrate and competes for the active site. The effect can be overcome by increasing $[S]$. $V_{\max}$ is unchanged but $K_m$ increases.

Non-competitive inhibition: the inhibitor binds to an allosteric site (not the active site), changing the enzyme's conformation and reducing its activity. This cannot be overcome by increasing $[S]$. $V_{\max}$ decreases but $K_m$ is unchanged.

Feature	Competitive	Non-competitive
Binds to	Active site	Allosteric site
Effect of increasing $[S]$	Overcomes inhibition	Cannot overcome
$V_{\max}$	Unchanged	Decreased
$K_m$	Increased	Unchanged
Example	Malonate inhibiting succinate dehydrogenase	Lead, cyanide, heavy metals

6.6 Immobilised Enzymes

Enzymes can be immobilised by binding them to an inert matrix (alginate beads, agar gel, membranes). Advantages:

• Enzymes can be recovered and reused
• Products are not contaminated with enzyme
• Enzymes are more stable at extreme temperatures and pH
• Multiple enzymes can be arranged in sequence for metabolic pathways

Disadvantages: reduced activity due to restricted substrate access; immobilisation is costly.

7. Quantitative Biochemistry: Food Tests and Standard Curves

7.1 Benedict's Test: Quantitative Analysis

The Benedict's test can be made semi-quantitative by comparing the colour of the precipitate against a standard colour chart, or by filtering the precipitate and weighing it. Alternatively, a colorimeter can be used to measure the absorbance of the solution at a specific wavelength (e.g., $630\ \mathrm{nm}$), and a standard curve constructed.

Worked Example: Constructing a Standard Curve. A series of glucose standards of known concentration is prepared and tested with Benedict's reagent. The absorbance of each is measured with a colorimeter:

Glucose concentration ($\mathrm{mg\ cm^{-3}}$)	0	2	4	6	8	10
Absorbance (arbitrary units)	0.0	0.2	0.4	0.6	0.8	1.0

The standard curve is linear: absorbance $= 0.1 \times$ concentration. If an unknown solution gives an absorbance of $0.65$, its glucose concentration is $0.65 / 0.1 = 6.5\ \mathrm{mg\ cm^{-3}}$.

7.2 Biuret Test: Quantitative Analysis

The intensity of the violet/purple colour in the Biuret test is proportional to the concentration of peptide bonds. This relationship can be used to determine the concentration of protein in an unknown solution using a standard curve prepared from protein solutions of known concentration (e.g., bovine serum albumin, BSA).

Worked Example. A set of BSA standards gives the following absorbance readings at $540\ \mathrm{nm}$:

Protein concentration ($\mathrm{mg\ cm^{-3}}$)	0	1	2	3	4	5
Absorbance	0.0	0.12	0.25	0.36	0.49	0.61

A student plots absorbance against concentration and determines the gradient (by line of best fit) to be approximately $0.122\ \mathrm{cm^3\ mg^{-1}}$. An unknown protein solution gives absorbance $= 0.42$.

Concentration $= 0.42 / 0.122 = 3.44\ \mathrm{mg\ cm^{-3}}$.

warning

Common Pitfall The Biuret test detects peptide bonds, not specific proteins. The absorbance reading gives total protein concentration regardless of protein type. Additionally, free amino acids do not give a reliable result because they lack peptide bonds. Ensure the standard curve is plotted correctly with absorbance on the y-axis and concentration on the x-axis.

8. Detailed Protein Structure: Case Studies

8.1 Haemoglobin as a Quaternary Protein

Haemoglobin (Hb) is a globular quaternary protein with four polypeptide chains: two $\alpha$-chains (141 amino acids each) and two $\beta$-chains (146 amino acids each). Each chain contains a haem group (protoporphyrin IX ring with a central $\mathrm{Fe^{2+}}$ ion) that binds one $\mathrm{O_2}$ molecule.

The quaternary structure is essential for function: the cooperative binding of $\mathrm{O_2}$ arises from conformational changes transmitted between subunits. When one haem group binds $\mathrm{O_2}$, the iron ion moves into the plane of the porphyrin ring, pulling the attached histidine residue and shifting the position of the entire subunit. This conformational change increases the affinity of the remaining three haem groups for $\mathrm{O_2}$.

For further detail on the oxygen dissociation curve and the Bohr effect, see Exchange and Transport.

8.2 Collagen as a Fibrous Protein

Collagen is the most abundant protein in the human body and exemplifies fibrous protein structure. Its properties derive directly from its structure at every level:

1. Primary structure: a repeating sequence of Gly-Pro-X (where X is often hydroxyproline). Glycine (the smallest amino acid, with $R = -\mathrm{H}$) is essential because the three polypeptide chains pack very tightly in the triple helix, and only glycine is small enough to fit at the centre.

2. Secondary structure: each chain forms a left-handed helix (not an $\alpha$-helix), stabilised by hydrogen bonds.

3. Tertiary structure: three polypeptide chains wind around each other to form a right-handed triple helix (supercoil), stabilised by hydrogen bonds between the chains and covalent cross-links between lysine residues.

4. Quaternary structure: triple helices align parallel and are cross-linked to form collagen fibrils, which bundle into collagen fibres. The cross-links provide tensile strength.

The relationship between collagen structure and function is a classic examination topic:

Structural Feature	Functional Consequence
Triple helix	High tensile strength; resists stretching forces
Regular repeating sequence	Uniform, ordered structure
Covalent cross-links	Fibrils are very strong and insoluble
Parallel alignment	Strength is directional (along the fibre axis)
Glycine at every third position	Allows tight packing of the three chains

warning

Common Pitfall Students often describe collagen as having a "globular" structure or describe its helix as an $\alpha$-helix. Collagen is a fibrous protein with a triple helix, which is structurally distinct from the $\alpha$-helix found in globular proteins like haemoglobin.

8.3 Sickle Cell Anaemia: A Point Mutation with Structural Consequences

Sickle cell anaemia is caused by a single base substitution (missense mutation) in the gene for the $\beta$-chain of haemoglobin: the codon GAG (glutamic acid) is changed to GTG (valine) at position 6.

This single amino acid change has profound consequences:

• Glutamic acid is hydrophilic (polar, charged). Valine is hydrophobic (non-polar).
• The hydrophobic valine on the surface of deoxygenated HbS (sickle haemoglobin) interacts with hydrophobic patches on adjacent HbS molecules.
• This causes HbS to polymerise into long, rigid fibres that distort the red blood cell into a characteristic sickle shape.
• Sickled cells are less flexible, block capillaries, and are destroyed prematurely (haemolytic anaemia).

This case study illustrates the central dogma of molecular biology: DNA sequence determines amino acid sequence, which determines protein structure, which determines protein function. A single base change in DNA propagates through all levels to produce a severe disease phenotype.

9. Enzyme Kinetics in Depth

9.1 Lineweaver-Burk Plots

The Michaelis-Menten equation can be linearised by taking the reciprocal of both sides:

$\frac{1}{v} = \frac◆LB◆K_m◆RB◆◆LB◆V_{\max}◆RB◆ \cdot \frac{1}{[S]} + \frac◆LB◆1◆RB◆◆LB◆V_{\max}◆RB◆$

A plot of $\frac{1}{v}$ versus $\frac{1}{[S]}$ yields a straight line with:

• Y-intercept $= \frac◆LB◆1◆RB◆◆LB◆V_{\max}◆RB◆$
• X-intercept $= -\frac{1}{K_m}$
• Gradient $= \frac◆LB◆K_m◆RB◆◆LB◆V_{\max}◆RB◆$

Worked Example. An enzyme has $V_{\max} = 50\ \mu\mathrm{mol\ min^{-1}}$ and $K_m = 5\ \mathrm{mmol\ dm^{-3}}$. Calculate the reaction rate when $[S] = 2\ \mathrm{mmol\ dm^{-3}}$ using the Michaelis-Menten equation, and verify using the Lineweaver-Burk plot.

Using the Michaelis-Menten equation:

$v = \frac◆LB◆V_{\max}[S]◆RB◆◆LB◆K_m + [S]◆RB◆ = \frac◆LB◆50 \times 2◆RB◆◆LB◆5 + 2◆RB◆ = \frac{100}{7} = 14.3\ \mu\mathrm{mol\ min^{-1}}$

Using the Lineweaver-Burk form:

$\frac{1}{v} = \frac◆LB◆K_m◆RB◆◆LB◆V_{\max}◆RB◆ \cdot \frac{1}{[S]} + \frac◆LB◆1◆RB◆◆LB◆V_{\max}◆RB◆ = \frac{5}{50} \times \frac{1}{2} + \frac{1}{50} = 0.05 + 0.02 = 0.07$

$v = \frac{1}{0.07} = 14.3\ \mu\mathrm{mol\ min^{-1}}$

Both methods agree, as expected.

9.2 Effect of Temperature on Enzyme Rate: The $Q_{10}$ Coefficient

The $Q_{10}$ temperature coefficient quantifies the effect of temperature on reaction rate:

$Q_{10} = \left(\frac{v_2}{v_1}\right)^{\frac{10}{T_2 - T_1}}$

or more commonly, where $T_2 - T_1 = 10\ ^\circ\mathrm{C}$:

$Q_{10} = \frac{v_{(T+10)}}{v_T}$

Most biological reactions have $Q_{10} \approx 2$--$3$, meaning the rate doubles or triples for each $10\ ^\circ\mathrm{C}$ increase in temperature (within the range before denaturation).

Worked Example. An enzyme-catalysed reaction proceeds at a rate of $12\ \mu\mathrm{mol\ min^{-1}}$ at $20\ ^\circ\mathrm{C}$. If $Q_{10} = 2.5$, what is the expected rate at $30\ ^\circ\mathrm{C}$?

$v_{30} = Q_{10} \times v_{20} = 2.5 \times 12 = 30\ \mu\mathrm{mol\ min^{-1}}$

At $40\ ^\circ\mathrm{C}$:

$v_{40} = Q_{10} \times v_{30} = 2.5 \times 30 = 75\ \mu\mathrm{mol\ min^{-1}}$

However, this calculation assumes the enzyme has not denatured. If the optimum temperature is $37\ ^\circ\mathrm{C}$, the actual rate at $40\ ^\circ\mathrm{C}$ may be lower than predicted due to partial denaturation. The $Q_{10}$ relationship is only valid within the range where the enzyme is fully functional.

9.3 Practical Investigation: Effect of pH on Enzyme Activity

A common practical investigation involves measuring the rate of an enzyme-catalysed reaction at different pH values. A reliable method:

1. Prepare buffered solutions at a range of pH values (e.g., pH 3, 5, 7, 9, 11).
2. Add a fixed volume of enzyme solution to each buffer.
3. Start the reaction by adding a fixed volume of substrate solution.
4. Measure the rate of reaction (e.g., by measuring the volume of gas produced per minute, or by taking samples at timed intervals and stopping the reaction).
5. Plot rate versus pH.

The resulting graph is a bell curve with a peak at the optimum pH. The shape of the curve reflects the ionisation states of amino acid residues in the active site and the protein's tertiary structure.

warning

Common Pitfall In enzyme practicals, students often fail to control variables: temperature must be kept constant (using a water bath), the same enzyme concentration must be used in each trial, and timing must start the instant the enzyme and substrate are mixed. Failure to control these variables invalidates the experiment.

10. Inorganic Ions in Biology

10.1 Key Ions and Their Roles

Ion	Role
$\mathrm{Fe^{2+}}$	Component of haem group in haemoglobin; essential for $\mathrm{O_2}$ transport
$\mathrm{Na^+}$	Co-transport of glucose in ileum; nerve impulse transmission (action potential)
$\mathrm{K^+}$	Co-transport; maintaining resting membrane potential; stomatal opening
$\mathrm{Ca^{2+}}$	Bone and teeth structure; blood clotting cascade; muscle contraction (binds troponin)
$\mathrm{Mg^{2+}}$	Component of chlorophyll; co-factor for many enzymes (e.g., DNA polymerase, Rubisco)
$\mathrm{PO_4^{3-}}$	Component of ATP, DNA, RNA, phospholipids
$\mathrm{NO_3^-}$	Nitrogen source for amino acid synthesis
$\mathrm{H^+}$ and $\mathrm{OH^-}$	Determine pH; affect enzyme activity; proton gradients in respiration and photosynthesis

10.2 Hydrogen Ions and pH

The pH scale is defined as:

$\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$

where $[\mathrm{H^+}]$ is the hydrogen ion concentration in $\mathrm{mol\ dm^{-3}}$.

• Pure water: $[\mathrm{H^+}] = 10^{-7}\ \mathrm{mol\ dm^{-3}}$, so $\mathrm{pH} = 7$.
• Stomach acid: $[\mathrm{H^+}] \approx 0.01\ \mathrm{mol\ dm^{-3}}$, so $\mathrm{pH} \approx 2$.
• Blood: $[\mathrm{H^+}] \approx 4 \times 10^{-8}\ \mathrm{mol\ dm^{-3}}$, so $\mathrm{pH} \approx 7.4$.

A change of one pH unit represents a tenfold change in $[\mathrm{H^+}]$. Blood pH is tightly regulated between 7.35 and 7.45 by the carbonic acid-bicarbonate buffer system (see Exchange and Transport for the Bohr effect).

Practice Problems

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Problem 1

Explain how the structure of cellulose relates to its function in plant cell walls. Include reference to the type of glycosidic bond, hydrogen bonding, and the arrangement of microfibrils.

Answer. Cellulose is composed of $\beta$-glucose monomers joined by $\beta$-1,4 glycosidic bonds. This configuration causes each successive glucose molecule to be rotated $180^\circ$ relative to its neighbour, creating a linear, unbranched chain. Adjacent cellulose chains align parallel and form extensive hydrogen bonds between $-\mathrm{OH}$ groups on adjacent chains. These hydrogen-bonded chains bundle together into microfibrils, which are further bundled into macrofibrils. This creates a rigid, strong meshwork that resists tensile forces and prevents the cell from bursting under osmotic pressure. The $\beta$-1,4 bond also makes cellulose resistant to hydrolysis by most enzymes -- only cellulase (produced by some bacteria and fungi) can break it.

If you get this wrong, revise: Polysaccharides

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Problem 2

A student performs the Benedict's test on a solution and observes a green colour change. They then acid-hydrolyse a second sample, neutralise it, and repeat the test, observing a brick-red precipitate. Explain these results.

Answer. The initial green colour indicates a low concentration of reducing sugars. After acid hydrolysis and neutralisation, the brick-red precipitate indicates a high concentration of reducing sugars. This means the original sample contained a non-reducing sugar (such as sucrose). The acid hydrolysis broke the glycosidic bond, releasing the component monosaccharides (glucose and fructose), which are reducing sugars. The low initial reading suggests very little free reducing sugar was present; most carbohydrate was in the non-reducing disaccharide form.

If you get this wrong, revise: Testing for Carbohydrates

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Problem 3

Describe and explain the effect of increasing substrate concentration on the rate of an enzyme-catalysed reaction in the presence of (a) a competitive inhibitor and (b) a non-competitive inhibitor. Sketch both graphs.

Answer. (a) With a competitive inhibitor, the initial rate is lower at every $[S]$ compared to the uninhibited reaction. However, as $[S]$ increases, the substrate outcompetes the inhibitor for active sites, so the curve approaches the same $V_{\max}$ as the uninhibited reaction. $K_m$ is increased because a higher $[S]$ is needed to reach half $V_{\max}$. (b) With a non-competitive inhibitor, the rate is lower at all $[S]$, and the curve plateaus at a lower $V_{\max}$ regardless of how high $[S]$ becomes. $K_m$ is unchanged because the inhibitor does not affect the enzyme's affinity for substrate -- it simply reduces the number of functional enzyme molecules.

If you get this wrong, revise: Enzyme Inhibition

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Problem 4

Compare and contrast the structure and function of starch and cellulose. (6 marks)

Answer. Similarities: both are polysaccharides of glucose; both are joined by glycosidic bonds; both are formed by condensation reactions; both are found in plants. Differences: starch is made of $\alpha$-glucose with $\alpha$-1,4 (and some $\alpha$-1,6) glycosidic bonds, whereas cellulose is made of $\beta$-glucose with $\beta$-1,4 glycosidic bonds. Starch is a coiled, compact molecule that is soluble and functions as an energy store. Cellulose has straight, unbranched chains that form hydrogen-bonded microfibrils, making it insoluble and functioning as a structural material. Starch can be hydrolysed by amylase; cellulose cannot (except by cellulase).

If you get this wrong, revise: Polysaccharides

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Problem 5

A researcher measures the initial rate of an enzyme-catalysed reaction at different substrate concentrations, with and without a non-competitive inhibitor. The results are:

$[S]$ ($\mathrm{mmol\ dm^{-3}}$)	Rate without inhibitor ($\mu\mathrm{mol\ min^{-1}}$)	Rate with inhibitor ($\mu\mathrm{mol\ min^{-1}}$)
0.5	8	4
1.0	14	7
2.0	22	11
5.0	30	15
10.0	32	16

(a) Estimate $V_{\max}$ and $K_m$ for both conditions. (b) Explain why $K_m$ is the same in both cases.

Answer. (a) Without inhibitor: $V_{\max} \approx 33\ \mu\mathrm{mol\ min^{-1}}$ (rate approaches this value asymptotically). $K_m \approx 1.7\ \mathrm{mmol\ dm^{-3}}$ (the $[S]$ at which rate $= 16.5$). With inhibitor: $V_{\max} \approx 17\ \mu\mathrm{mol\ min^{-1}}$ (half the uninhibited value, consistent with the inhibitor reducing active enzyme concentration by 50%). $K_m \approx 1.7\ \mathrm{mmol\ dm^{-3}}$ (unchanged). (b) The non-competitive inhibitor binds to an allosteric site and does not affect the active site itself, so the enzyme's affinity for its substrate (reflected by $K_m$) is unchanged. It simply reduces the proportion of functional enzyme molecules available.

If you get this wrong, revise: Enzyme Kinetics

Details

Problem 6

Explain how the structure of a phospholipid molecule enables the spontaneous formation of cell membranes in aqueous environments. In your answer, refer to the hydrophobic effect.

Answer. Phospholipids are amphipathic: the phosphate head is hydrophilic (charged, forms hydrogen bonds with water), while the two fatty acid tails are hydrophobic (non-polar, cannot form favourable interactions with water). In aqueous solution, the second law of thermodynamics drives the system to maximise entropy. Water molecules form ordered cages (clathrate structures) around hydrophobic tails, which is entropically unfavourable. To minimise this, phospholipids spontaneously arrange so that hydrophobic tails are shielded from water while hydrophilic heads remain exposed. The most stable arrangement at the concentrations found in cells is a bilayer: two layers of phospholipids with tails facing inward and heads facing outward. This process is spontaneous ($\Delta G \lt 0$) because the increase in entropy of water molecules outweighs the decrease in entropy of the phospholipids themselves.

If you get this wrong, revise: Phospholipids

Details

Problem 7

The enzyme catalase breaks down hydrogen peroxide: $2\mathrm{H_2O_2} \to 2\mathrm{H_2O} + \mathrm{O_2}$. A student investigates the effect of catalase concentration on the rate of reaction by measuring the volume of $\mathrm{O_2}$ produced in 60 seconds. The results are:

Catalase concentration (arbitrary units)	Volume of $\mathrm{O_2}$ ($\mathrm{cm^3}$)
1	4.2
2	8.3
4	16.1
8	24.8
16	27.5

(a) Plot a graph of rate versus enzyme concentration. (b) Explain the shape of the graph. (c) At what concentration does the reaction become limited by substrate availability?

Answer. (b) At low catalase concentrations (1--4 units), the rate is approximately proportional to enzyme concentration (first-order kinetics: doubling enzyme concentration approximately doubles the rate). At higher concentrations (8--16 units), the rate begins to plateau (approaches zero-order kinetics). This is because the substrate ($\mathrm{H_2O_2}$) is becoming the limiting factor: at high enzyme concentration, substrate molecules are converted to product as fast as they encounter an enzyme, and additional enzyme molecules have nothing to process.

(c) The plateau begins between 8 and 16 units. By 16 units, increasing enzyme concentration from 8 to 16 (approximately doubling) only increases the rate from 24.8 to 27.5 ($\approx 11\%$ increase), indicating the reaction has nearly reached $V_{\max}$. Substrate is the limiting factor at concentrations above approximately 8 units.

If you get this wrong, revise: Enzyme Kinetics

Details

Problem 8

Explain why collagen is a structural protein whereas haemoglobin is a globular protein. Refer to the amino acid sequences, bonding, and how each protein's structure relates to its function. (6 marks)

Answer. Collagen has a repetitive primary structure (Gly-Pro-X) with glycine at every third position, allowing three chains to pack tightly into a triple helix. It is stabilised by extensive hydrogen bonding between chains and covalent cross-links, making it strong, insoluble, and rigid. These properties are suited to its structural role (connective tissue, tendons, skin). Haemoglobin has a varied primary structure with both hydrophilic and hydrophobic amino acids. The hydrophobic residues cluster in the interior and hydrophilic residues on the surface, making it soluble in water (globular). Its quaternary structure (four subunits) allows conformational changes between subunits, enabling cooperative $\mathrm{O_2}$ binding -- suited to its transport role. The key distinction is that structural proteins require insolubility and mechanical strength, whereas transport proteins require solubility and the ability to change conformation.

If you get this wrong, revise: Protein Types and Detailed Protein Structure: Case Studies

Details

Problem 9

An enzyme has $V_{\max} = 40\ \mu\mathrm{mol\ min^{-1}}$ and $K_m = 8\ \mathrm{mmol\ dm^{-3}}$. A competitive inhibitor is added at a concentration that doubles the apparent $K_m$. (a) What is the new $K_m$? (b) At $[S] = 8\ \mathrm{mmol\ dm^{-3}}$, calculate the rate with and without the inhibitor. (c) At $[S] = 100\ \mathrm{mmol\ dm^{-3}}$, are the rates still different?

Answer. (a) New $K_m = 2 \times 8 = 16\ \mathrm{mmol\ dm^{-3}}$. $V_{\max}$ is unchanged at $40\ \mu\mathrm{mol\ min^{-1}}$.

(b) Without inhibitor: $v = \frac◆LB◆40 \times 8◆RB◆◆LB◆8 + 8◆RB◆ = \frac{320}{16} = 20\ \mu\mathrm{mol\ min^{-1}}$.

With inhibitor: $v = \frac◆LB◆40 \times 8◆RB◆◆LB◆16 + 8◆RB◆ = \frac{320}{24} = 13.3\ \mu\mathrm{mol\ min^{-1}}$.

The inhibitor reduces the rate by 33% at this substrate concentration.

(c) Without inhibitor: $v = \frac◆LB◆40 \times 100◆RB◆◆LB◆8 + 100◆RB◆ = \frac{4000}{108} = 37.0\ \mu\mathrm{mol\ min^{-1}}$.

With inhibitor: $v = \frac◆LB◆40 \times 100◆RB◆◆LB◆16 + 100◆RB◆ = \frac{4000}{116} = 34.5\ \mu\mathrm{mol\ min^{-1}}$.

At high $[S]$, the rates converge towards $V_{\max}$, but are still slightly different because $[S]$ is not infinitely greater than $K_m$. As $[S] \to \infty$, both rates approach $V_{\max} = 40$.

If you get this wrong, revise: Enzyme Inhibition and Enzyme Kinetics in Depth

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11. Advanced Carbohydrate Chemistry

11.1 Alpha vs Beta Anomers

In aqueous solution, glucose exists in equilibrium between the linear (open-chain) form and two cyclic forms (alpha and beta anomers). The cyclisation occurs when the $\mathrm{C1}$ aldehyde group reacts with the $\mathrm{C5}$ hydroxyl group to form a 6-membered pyranose ring.

• Alpha-glucose: the $-\mathrm{OH}$ on $\mathrm{C1}$ is below the plane of the ring (trans to the $\mathrm{C6}\ \mathrm{-CH_2OH}$ group).
• Beta-glucose: the $-\mathrm{OH}$ on $\mathrm{C1}$ is above the plane of the ring (cis to the $\mathrm{C6}\ \mathrm{-CH_2OH}$ group).

This difference appears minor (one $-\mathrm{OH}$ group flipped) but has enormous structural consequences:

• Amylose (starch) is made of $\alpha$-glucose joined by $\alpha$-1,4-glycosidic bonds. The chain coils into a helix, making it compact and suitable for energy storage.
• Cellulose is made of $\beta$-glucose joined by $\beta$-1,4-glycosidic bonds. Every alternate glucose molecule is rotated $180$ degrees, so the $-\mathrm{OH}$ groups project from opposite sides of the chain. This allows extensive hydrogen bonding between adjacent chains, forming rigid, insoluble fibres ideal for structural support.

11.2 Carbohydrate Digestion and Absorption

Digestion of starch:

1. Salivary amylase (in the mouth, pH $\approx 6.8$) hydrolyses $\alpha$-1,4-glycosidic bonds in amylose and amylopectin, producing maltose, maltotriose, and $\alpha$-limit dextrins (from amylopectin's $\alpha$-1,6 branches).
2. Pancreatic amylase (in the duodenum, pH $\approx 7.5$) continues the same hydrolysis.
3. Brush border enzymes on the microvilli of the ileum epithelial cells complete digestion:
   • Maltase: maltose $\to$ glucose + glucose.
   • Isomaltase: $\alpha$-1,6 bonds in limit dextrins $\to$ glucose.
   • Sucrase: sucrose $\to$ glucose + fructose.
   • Lactase: lactose $\to$ glucose + galactose.

Absorption: glucose and galactose are absorbed by secondary active transport via the SGLT1 co-transporter ($\mathrm{Na^+}$-glucose linked transporter), which uses the $\mathrm{Na^+}$ gradient maintained by the $\mathrm{Na^+/K^+}$ ATPase pump. Fructose is absorbed by facilitated diffusion via the GLUT5 transporter. All three monosaccharides exit the cell into the bloodstream via GLUT2.

11.3 Lactose Intolerance

Lactose intolerance results from insufficient lactase production in adulthood (primary lactase deficiency). Undigested lactose passes to the large intestine, where it is fermented by gut bacteria, producing gas ($\mathrm{CO_2}$, $\mathrm{H_2}$, methane) and organic acids. This causes bloating, flatulence, abdominal cramps, and osmotic diarrhoea (the undigested lactose increases the osmolarity of the intestinal contents, drawing water into the lumen).

Lactase persistence in adults is caused by mutations upstream of the LCT gene on chromosome 2, which allow continued lactase expression. This trait evolved independently in European, East African, and Middle Eastern populations -- a classic example of convergent evolution driven by the cultural practice of dairy farming (gene-culture coevolution).

12. Advanced Lipid Chemistry

12.1 Phospholipid Structure and Membrane Properties

Phospholipids are amphipathic: they have a hydrophilic (polar) head (phosphate group + organic base such as choline, serine, or ethanolamine) and a hydrophobic (non-polar) tail (two fatty acid chains).

The degree of saturation of the fatty acid tails affects membrane fluidity:

• Saturated fatty acids (no double bonds, e.g., palmitic acid $\mathrm{C_{16}:0}$): straight chains pack tightly, reducing membrane fluidity.
• Unsaturated fatty acids (one or more cis double bonds, e.g., oleic acid $\mathrm{C_{18}:1}$): kinked chains prevent tight packing, increasing membrane fluidity.

Organisms regulate membrane fluidity in response to temperature:

• At low temperature, they increase the proportion of unsaturated fatty acids (more kinks = more fluid).
• At high temperature, they increase the proportion of saturated fatty acids (straighter chains = less fluid).

12.2 Cholesterol in Membranes

Cholesterol is a sterol lipid that is embedded in animal cell membranes, with its small hydroxyl group at the hydrophilic head region and its rigid steroid ring structure within the hydrophobic core.

Cholesterol acts as a fluidity buffer:

• At high temperature, it reduces membrane fluidity by restricting the movement of phospholipid fatty acid chains.
• At low temperature, it prevents tight packing by inserting between phospholipid molecules, maintaining fluidity.

Cholesterol also reduces the permeability of the membrane to small water-soluble molecules and ions.

12.3 Lipid Digestion, Absorption, and Transport

Digestion:

1. Bile salts (produced in the liver, stored in the gall bladder) emulsify fats in the duodenum: they break large fat globules into smaller droplets (micelles), increasing the surface area for enzyme action.
2. Pancreatic lipase hydrolyses triglycerides into monoglycerides and fatty acids.
3. Cholesterol esterase hydrolyses cholesterol esters into free cholesterol and fatty acids.

Absorption: monoglycerides, fatty acids, and cholesterol are absorbed into the epithelial cells of the ileum by simple diffusion (they are non-polar and can cross the phospholipid bilayer). Inside the cell, they are re-esterified and packaged into chylomicrons (lipoprotein particles coated with phospholipids and proteins).

Transport: chylomicrons are too large to enter blood capillaries, so they enter the lacteals (lymphatic capillaries in the villi) and are carried in the lymphatic system to the thoracic duct, which empties into the bloodstream near the heart.

12.4 Atherosclerosis

Atherosclerosis is the buildup of fatty plaques (atheromas) in the walls of arteries. The process:

1. Endothelial damage (by high blood pressure, smoking, or high LDL cholesterol) allows LDL to infiltrate the arterial wall.
2. LDL oxidation triggers an inflammatory response: monocytes differentiate into macrophages that engulf oxidised LDL, becoming foam cells.
3. Foam cells accumulate, forming a fatty streak.
4. A fibrous cap of smooth muscle cells and collagen forms over the fatty deposit.
5. The plaque narrows the artery lumen (stenosis), reducing blood flow.
6. If the fibrous cap ruptures, a blood clot (thrombus) forms, which can block the artery entirely, causing a myocardial infarction (heart attack) or stroke.

13. Advanced Protein Chemistry

13.1 Protein Folding and Chaperones

Newly synthesised polypeptides must fold into their correct tertiary structure to become functional. Molecular chaperones (chaperonins) assist folding:

1. The unfolded polypeptide enters the chaperonin cavity (e.g., GroEL/GroES in bacteria, TRiC in eukaryotes).
2. ATP hydrolysis changes the chaperonin's conformation, providing an isolated environment for folding.
3. After a set time, the polypeptide is released. If it is not correctly folded, it re-enters for another round.

Misfolded proteins are targeted for degradation by the ubiquitin-proteasome system: ubiquitin (a small protein) is attached to the misfolded protein, which is then recognised and degraded by the proteasome (a barrel-shaped protease complex).

13.2 Prions

Prions ($\mathrm{PrP^{Sc}}$) are infectious protein particles that cause neurodegenerative diseases (e.g., Creutzfeldt-Jakob disease in humans, BSE in cattle, scrapie in sheep).

The normal cellular prion protein ($\mathrm{PrP^C}$) is a $\alpha$-helix-rich protein on the surface of neurons. The infectious form ($\mathrm{PrP^{Sc}}$) has a different conformation with more $\beta$-sheet content. When $\mathrm{PrP^{Sc}}$ contacts $\mathrm{PrP^C}$, it converts it to the abnormal conformation -- a chain reaction.

Key features of prions:

• They contain no nucleic acid (no DNA or RNA) -- the information is carried purely in the protein's 3D structure.
• They are extremely resistant to heat (survive autoclaving at $121$ degrees C), UV radiation, and proteases.
• They cause irreversible brain damage: $\mathrm{PrP^{Sc}}$ aggregates into amyloid plaques that damage neurons, causing sponge-like holes in the brain tissue (spongiform encephalopathy).

14. Nucleic Acid Chemistry in Depth

14.1 DNA Supercoiling

In vivo, DNA is supercoiled: the double helix itself is coiled around histone proteins to form nucleosomes. Supercoiling allows the long DNA molecule (approximately 2 m per human cell) to fit inside the nucleus (diameter $\approx 5\ \mu\mathrm{m}$).

• Negative supercoiling (underwinding) makes it easier to separate the strands for transcription and replication.
• Positive supercoiling (overwinding) occurs ahead of the replication fork and is relieved by topoisomerase enzymes (e.g., DNA gyrase in bacteria), which cut one or both strands, allow rotation, and reseal them.

14.2 RNA Types and Functions

RNA Type	Full Name	Structure	Function
mRNA	Messenger RNA	Single-stranded, with 5' cap and 3' poly-A tail	Carries the genetic code from DNA to the ribosome
tRNA	Transfer RNA	Cloverleaf secondary structure; L-shaped tertiary structure; 3' CCA end	Carries specific amino acids to the ribosome; anticodon pairs with mRNA codon
rRNA	Ribosomal RNA	Large, complex secondary and tertiary structure	Catalytic and structural component of ribosomes (rRNA is the ribozyme, not protein)
snRNA	Small nuclear RNA	Short, complexed with proteins to form snRNPs	Splicing of pre-mRNA (removal of introns) in the spliceosome
miRNA	Micro RNA	Short ($\approx 22\ \mathrm{nt}$), single-stranded	Post-transcriptional gene regulation: binds to mRNA and inhibits translation or promotes degradation
siRNA	Small interfering RNA	Short double-stranded RNA	RNA interference: triggers degradation of complementary mRNA

14.3 Telomeres and the Hayflick Limit

Telomeres are repetitive DNA sequences (TTAGGG in vertebrates, repeated hundreds to thousands of times) at the ends of chromosomes. They protect the coding regions of chromosomes from degradation and prevent chromosomes from fusing with each other.

During DNA replication, the leading strand is synthesised completely, but the lagging strand has a problem: the RNA primer at the very end of the chromosome cannot be replaced with DNA (there is no upstream $3'$-OH for DNA polymerase to extend from). This means approximately 50--200 base pairs are lost from the telomere with each cell division.

After approximately 40--60 cell divisions (the Hayflick limit), the telomere becomes critically short, triggering cellular senescence (permanent cell cycle arrest) or apoptosis (programmed cell death). This is thought to contribute to ageing.

Telomerase is an enzyme (a reverse transcriptase with an RNA template) that can extend telomeres by adding TTAGGG repeats. It is active in germ cells, stem cells, and most cancer cells, allowing them to divide indefinitely. Telomerase is inactive in most somatic cells.

15. Inorganic Ions in Depth

15.1 Calcium Ions ($\mathrm{Ca^{2+}}$)

Calcium ions are essential second messengers in cell signalling. They are stored at high concentration in the sarcoplasmic reticulum (muscle cells) and endoplasmic reticulum (other cells), and at low concentration in the cytoplasm ($\approx 0.1\ \mu\mathrm{M}$).

Role in muscle contraction: when an action potential arrives at a neuromuscular junction, acetylcholine is released, triggering an action potential in the muscle fibre. This depolarisation causes $\mathrm{Ca^{2+}}$ release from the sarcoplasmic reticulum. $\mathrm{Ca^{2+}}$ binds to troponin C, causing a conformational change that moves tropomyosin away from the myosin-binding sites on actin filaments. Myosin heads can then bind to actin and undergo the power stroke, causing muscle contraction.

Role in blood clotting: $\mathrm{Ca^{2+}}$ (Factor IV) is essential at multiple steps in the clotting cascade. It is required for the activation of several clotting factors (Factors II, VII, IX, X) and for the conversion of prothrombin to thrombin. Anticoagulants such as EDTA (ethylenediaminetetraacetic acid) chelate $\mathrm{Ca^{2+}}$, preventing clotting in blood samples.

15.2 Iron Ions ($\mathrm{Fe^{2+/3+}}$)

Iron is a component of haem, the prosthetic group in haemoglobin, myoglobin, and cytochromes. In haemoglobin, each of the four polypeptide chains contains one haem group, each of which binds one $\mathrm{O_2}$ molecule.

Iron exists in two oxidation states in biology:

• $\mathrm{Fe^{2+}}$ (ferrous): binds $\mathrm{O_2}$ in haemoglobin and myoglobin.
• $\mathrm{Fe^{3+}}$ (ferric): does not bind $\mathrm{O_2}$. Oxidation of $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ produces methaemoglobin, which cannot carry $\mathrm{O_2}$.

Iron deficiency causes anaemia (reduced haemoglobin, reduced $\mathrm{O_2}$-carrying capacity, fatigue, pallor). Iron overload (haemochromatosis) causes liver damage, diabetes, and heart failure.

15.3 Hydrogen Ions ($\mathrm{H^+}$) and Buffers

$\mathrm{H^+}$ concentration determines pH ($\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$). Biological systems require tight pH regulation because enzyme activity is pH-dependent.

Buffer systems resist pH changes:

1. Bicarbonate buffer (blood): $\mathrm{H^+ + HCO_3^- \rightleftharpoons H_2CO_3 \rightleftharpoons CO_2 + H_2O}$. This is the most important blood buffer because both components ($\mathrm{HCO_3^-}$ and $\mathrm{CO_2}$) can be regulated: $\mathrm{CO_2}$ by ventilation, $\mathrm{HCO_3^-}$ by the kidneys.
2. Phosphate buffer (intracellular, urine): $\mathrm{H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}}$, $\mathrm{p}K_a \approx 7.2$ (close to physiological pH).
3. Protein buffer (blood): haemoglobin binds $\mathrm{H^+}$ (forming haemoglobinic acid), buffering the $\mathrm{H^+}$ produced by $\mathrm{CO_2}$ transport (the Bohr effect).

warning

Common Pitfall Students often write that "buffers prevent pH change." This is incorrect. Buffers minimise pH change -- they do not prevent it entirely. A buffer resists pH change by providing a reservoir of both a weak acid and its conjugate base, which can react with added $\mathrm{H^+}$ or $\mathrm{OH^-}$.

19. Lipid Digestion and Transport in Detail

19.1 Emulsification

Bile salts (produced in the liver and stored in the gall bladder) emulsify fats into smaller droplets:

1. Bile salts are amphipathic molecules (have both hydrophilic and hydrophobic regions).
2. The hydrophilic regions interact with water, while the hydrophobic regions interact with the fat droplet.
3. This reduces the surface tension and breaks large fat globules into smaller droplets (1--2 $\mu\mathrm{m}$ diameter), increasing the surface area available for lipase action.

19.2 Pancreatic Lipase

Pancreatic lipase hydrolyses triglycerides at the oil-water interface:

$\text{Triglyceride} + 2\ \text{H}_2\text{O} \to \text{Monoacylglycerol} + 2\ \text{Fatty acids}$

The products (monoacylglycerol and fatty acids) are absorbed into the epithelial cells of the ileum by simple diffusion (they are non-polar and can cross the phospholipid bilayer).

19.3 Micelles and Chylomicron Formation

Inside the epithelial cell, fatty acids and monoacylglycerol are re-esterified into triglycerides. These are packaged with cholesterol and fat-soluble vitamins (A, D, E, K) into chylomicrons (lipoprotein particles with a protein coat).

Chylomicrons are too large to enter blood capillaries. They enter lacteals (lymphatic capillaries in the villi) and are transported via the lymphatic system to the bloodstream via the thoracic duct.

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tip

Diagnostic Test

17. Nucleic Acids: Advanced Topics

17.1 DNA Replication: Origins and Direction

Eukaryotic chromosomes have multiple origins of replication (approximately 10,000 per chromosome in humans), allowing replication to proceed simultaneously from many points. This reduces the time required to replicate the entire genome (approximately 8 hours in human cells, compared to approximately 40 minutes in E. coli which has a single origin and a much smaller genome).

Replication proceeds bidirectionally from each origin, producing replication bubbles that eventually merge as replication forks from adjacent origins meet.

17.2 RNA Processing: Detailed Steps

After transcription, the primary RNA transcript undergoes:

1. 5' capping: addition of 7-methylguanosine to the 5' end. This cap protects the mRNA from exonucleases and is recognised by the ribosome during translation initiation.

2. Splicing: introns are removed by the spliceosome (a complex of snRNPs). The splicing reaction involves two transesterification steps:

   • First, the 2'-OH of the branch point adenosine attacks the 5' splice site, cleaving the 5' exon and forming a lariat structure with the intron.
   • Second, the 3'-OH of the released 5' exon attacks the 3' splice site, joining the exons and releasing the intron lariat.

3. 3' polyadenylation: the primary transcript is cleaved downstream of the polyadenylation signal (AAUAAA), and approximately 200 adenine nucleotides are added by poly-A polymerase. The poly-A tail protects the mRNA from degradation and aids in its export from the nucleus.

17.3 Types of RNA Polymerase in Eukaryotes

Polymerase	Product	Location
RNA polymerase I	rRNA (except 5S rRNA)	Nucleolus
RNA polymerase II	mRNA, snRNA, microRNA	Nucleoplasm
RNA polymerase III	tRNA, 5S rRNA, other small RNAs	Nucleoplasm

17.4 Calculating mRNA Length from Gene Data

Worked Example. A gene has the following structure:

• 5' UTR: 150 nucleotides
• Exon 1: 450 nucleotides
• Intron 1: 1200 nucleotides
• Exon 2: 300 nucleotides
• Intron 2: 800 nucleotides
• Exon 3: 600 nucleotides
• 3' UTR: 200 nucleotides

Mature mRNA length (after splicing): $150 + 450 + 300 + 600 + 200 = 1700$ nucleotides.

Coding sequence (exons only, excluding UTRs): $450 + 300 + 600 = 1350$ nucleotides.

Number of amino acids in the protein: $\frac{1350}{3} = 450$ (minus 1 for the stop codon) $= 449$ amino acids.

17.5 Antisense RNA and RNA Interference

Antisense RNA is a single-stranded RNA molecule that is complementary to a specific mRNA molecule. When antisense RNA binds to its target mRNA, it forms a double-stranded RNA molecule that cannot be translated, effectively silencing the gene.

RNA interference (RNAi) is a natural regulatory mechanism:

1. Double-stranded RNA (dsRNA) is processed by the enzyme Dicer into short interfering RNAs (siRNAs), approximately 21--23 nucleotides long.
2. The siRNA is loaded into the RNA-induced silencing complex (RISC).
3. The guide strand of the siRNA directs RISC to complementary mRNA sequences.
4. RISC cleaves the target mRNA, preventing translation.

RNAi has been exploited as a research tool (gene knockdown) and has potential therapeutic applications (silencing disease-causing genes).

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tip

Diagnostic Test

16. Enzyme Inhibition: Extended Analysis

16.1 Lineweaver-Burk Plots

The Lineweaver-Burk plot is a double-reciprocal plot ($\frac{1}{v}$ vs $\frac{1}{[S]}$) that linearises the Michaelis-Menten equation:

$\frac{1}{v} = \frac◆LB◆K_m◆RB◆◆LB◆V_{\max}◆RB◆ \cdot \frac{1}{[S]} + \frac◆LB◆1◆RB◆◆LB◆V_{\max}◆RB◆$

The y-intercept $= \frac◆LB◆1◆RB◆◆LB◆V_{\max}◆RB◆$; the x-intercept $= -\frac{1}{K_m}$; the slope $= \frac◆LB◆K_m◆RB◆◆LB◆V_{\max}◆RB◆$.

Competitive inhibition: the slope increases, the y-intercept is unchanged (same $V_{\max}$), and the x-intercept moves closer to zero (higher apparent $K_m$).

Non-competitive inhibition: the y-intercept increases (lower $V_{\max}$), the slope increases, but the x-intercept is unchanged (same $K_m$).

16.2 Worked Example: Determining $K_m$ and $V_{\max}$

A student measures the rate of an enzyme-catalysed reaction at different substrate concentrations:

$[S]$ ($\mathrm{mM}$)	Rate $v$ ($\mu\mathrm{mol\ min^{-1}}$)	$\frac{1}{[S]}$ ($\mathrm{mM^{-1}}$)	$\frac{1}{v}$ ($\mathrm{min\ \mu mol^{-1}}$)
2	6.7	0.500	0.149
5	13.3	0.200	0.075
10	20.0	0.100	0.050
20	26.7	0.050	0.037
50	33.3	0.020	0.030
100	36.4	0.010	0.027

Plotting $\frac{1}{v}$ vs $\frac{1}{[S]}$:

At very high $[S]$: $v$ approaches $V_{\max} \approx 40\ \mu\mathrm{mol\ min^{-1}}$.

From the Michaelis-Menten equation: $v = \frac◆LB◆V_{\max} [S]◆RB◆◆LB◆K_m + [S]◆RB◆$.

At $v = \frac◆LB◆V_{\max}◆RB◆◆LB◆2◆RB◆$: $[S] = K_m$.

$\frac◆LB◆V_{\max}◆RB◆◆LB◆2◆RB◆ = 20\ \mu\mathrm{mol\ min^{-1}}$. At $v = 20$, $[S] = 10\ \mathrm{mM}$.

Therefore: $K_m = 10\ \mathrm{mM}$ and $V_{\max} = 40\ \mu\mathrm{mol\ min^{-1}}$.

20. Proteins: Advanced Structure and Function

20.1 Levels of Protein Structure

Level	Description	Bonds/Interactions	Example
Primary	Sequence of amino acids in the polypeptide chain	Peptide bonds (covalent)	Any protein (unique to each)
Secondary	Local folding into regular structures	Hydrogen bonds between backbone C=O and N-H groups	$\alpha$-helix (keratin, collagen-like triple helix); $\beta$-pleated sheet (silk fibroin)
Tertiary	Overall 3D shape of a single polypeptide	Hydrogen bonds, ionic bonds, disulphide bridges, hydrophobic interactions, van der Waals forces	Lysozyme, myoglobin
Quaternary	Assembly of multiple polypeptide subunits	Same as tertiary (between subunits)	Haemoglobin ($\alpha_2\beta_2$), immunoglobulin G (2 heavy + 2 light chains)

20.2 Haemoglobin and Oxygen Transport

Haemoglobin structure:

• Quaternary protein with 4 subunits (2 $\alpha$ chains, 2 $\beta$ chains).
• Each subunit contains a haem group (protoporphyrin ring with an $\mathrm{Fe^{2+}}$ ion at the centre).
• Each haem group can bind one $\mathrm{O_2}$ molecule, so each haemoglobin molecule can carry up to 4 $\mathrm{O_2}$ molecules.

Oxygen dissociation curves:

The oxyhaemoglobin dissociation curve is sigmoidal (S-shaped) because of cooperative binding: binding of the first $\mathrm{O_2}$ molecule induces a conformational change (T state $\to$ R state) that increases the affinity of the remaining haem groups for $\mathrm{O_2}$.

Factor	Effect on Curve	Explanation
High $\mathrm{pCO_2}$ (Bohr effect)	Right shift	$\mathrm{H^+}$ ions bind to haemoglobin, reducing its affinity for $\mathrm{O_2}$ (promoting $\mathrm{O_2}$ release in metabolically active tissues)
Low pH (high $[\mathrm{H^+}]$)	Right shift	Same mechanism as Bohr effect
High temperature	Right shift	Reduces haemoglobin's affinity for $\mathrm{O_2}$ (more $\mathrm{O_2}$ released in active muscles, which generate heat)
High 2,3-BPG concentration	Right shift	2,3-BPG binds to deoxyhaemoglobin, stabilising the T state and promoting $\mathrm{O_2}$ unloading
Foetal haemoglobin (HbF)	Left shift	Higher affinity for $\mathrm{O_2}$ than adult haemoglobin (HbA), allowing efficient $\mathrm{O_2}$ transfer from mother to foetus across the placenta

20.3 Enzyme Inhibition: Competitive vs Non-Competitive

Feature	Competitive Inhibition	Non-Competitive Inhibition
Inhibitor structure	Similar to substrate (structural analogue)	Not similar to substrate
Binding site	Active site	Allosteric site (different from active site)
Effect on $K_m$	Increases (apparent $K_m$ increases because higher $[S]$ is needed to achieve half $V_{\max}$)	No change
Effect on $V_{\max}$	No change (at very high $[S]$, substrate outcompetes inhibitor)	Decreases
Can be overcome?	Yes, by increasing substrate concentration	No, because inhibitor binds at a different site
Example	Malonate inhibits succinate dehydrogenase (Krebs cycle)	Heavy metals ($\mathrm{Pb^{2+}}$, $\mathrm{Hg^{2+}}$) bind to -SH groups, changing enzyme shape

20.4 Collagen: A Structural Protein

Collagen is the most abundant protein in the human body (approximately 25--35% of total body protein). Key features:

• Primary structure: repeating sequence Gly-X-Y (where X is often proline and Y is often hydroxyproline). Glycine (the smallest amino acid) is essential because it fits in the centre of the triple helix.
• Secondary structure: each chain forms a left-handed helix (not an $\alpha$-helix).
• Tertiary structure: three polypeptide chains wind around each other to form a right-handed triple helix (superhelix), stabilised by hydrogen bonds between the chains.
• Quaternary structure: triple helices are cross-linked by covalent bonds to form collagen fibrils, which bundle into collagen fibres.

Collagen provides tensile strength (resistance to stretching) in tendons, skin, bone, and connective tissue. Vitamin C is required as a cofactor for the enzyme prolyl hydroxylase, which hydroxylates proline to hydroxyproline. Hydroxyproline stabilises the triple helix by forming additional hydrogen bonds. Vitamin C deficiency causes scurvy (collagen fibres are weak, leading to bleeding gums, poor wound healing, and joint pain).

21. Nucleic Acids: Advanced Structure

21.1 DNA Double Helix: Key Features

• Two antiparallel polynucleotide strands wound around each other in a right-handed double helix.
• The sugar-phosphate backbone is on the outside; the nitrogenous bases are on the inside, stacked perpendicular to the axis of the helix.
• Base pairing: A pairs with T (2 hydrogen bonds); G pairs with C (3 hydrogen bonds). This is called complementary base pairing.
• The diameter of the helix is approximately 2 nm. One complete turn of the helix is approximately 3.4 nm and contains 10 base pairs.
• The double helix has a major groove and a minor groove, which are binding sites for DNA-binding proteins (e.g., transcription factors, histones).

21.2 DNA Supercoiling

DNA is supercoiled to allow it to fit inside the cell:

• Prokaryotes: circular DNA is negatively supercoiled (underwound) by DNA gyrase (topoisomerase II), which introduces negative supercoils by breaking both strands, passing one through the other, and resealing. Negative supercoiling makes it easier to separate the strands for transcription and replication.
• Eukaryotes: linear DNA is wrapped around histone proteins to form nucleosomes. Each nucleosome consists of DNA wrapped approximately 1.65 times around an octamer of histones (2 $\times$ H2A, 2 $\times$ H2B, 2 $\times$ H3, 2 $\times$ H4). Nucleosomes are connected by linker DNA (bound by histone H1) to form a "beads on a string" structure, which is further coiled into a solenoid (30 nm fibre), then looped and condensed into chromosomes.

21.3 Types of RNA

RNA Type	Full Name	Structure	Function
mRNA	Messenger RNA	Single-stranded; 5' cap and 3' poly-A tail; codons	Carries genetic information from DNA to ribosomes; template for translation
tRNA	Transfer RNA	Cloverleaf secondary structure; L-shaped tertiary structure; anticodon loop; 3' CCA end (amino acid attachment site)	Carries specific amino acids to the ribosome; matches anticodon to mRNA codon
rRNA	Ribosomal RNA	Single-stranded with extensive secondary structure; forms the core of ribosomes	Catalytic (rRNA is the ribozyme in the large subunit that forms peptide bonds); structural component of ribosomes
miRNA	Micro RNA	Short (~22 nucleotides); single-stranded; binds to complementary sequences in 3' UTR of mRNA	Gene regulation: inhibits translation or promotes mRNA degradation
snRNA	Small nuclear RNA	~150 nucleotides; associates with proteins to form snRNPs (small nuclear ribonucleoproteins)	Involved in mRNA splicing (removes introns from pre-mRNA)
siRNA	Small interfering RNA	~21--23 nucleotides; double-stranded	RNA interference: guides RISC to cleave complementary mRNA

21.4 DNA Replication: Leading and Lagging Strands

DNA replication is semi-conservative (each new DNA molecule consists of one original strand and one new strand) and semi-discontinuous (the leading strand is synthesised continuously, but the lagging strand is synthesised in short fragments called Okazaki fragments):

Feature	Leading Strand	Lagging Strand
Direction of synthesis	5' $\to$ 3' (same direction as replication fork movement)	5' $\to$ 3' (opposite direction to fork movement)
Synthesis	Continuous	Discontinuous (Okazaki fragments, 1000--2000 nucleotides in eukaryotes)
Primer requirement	One RNA primer at the origin of replication	Multiple RNA primers (one for each Okazaki fragment)
Enzyme	DNA polymerase III (prokaryotes) or DNA polymerases $\delta$ and $\varepsilon$ (eukaryotes)	DNA polymerase III or $\delta$; Okazaki fragments are later joined by DNA ligase

21.5 The Genetic Code

Feature	Description
Degenerate	Most amino acids are encoded by more than one codon (64 codons for 20 amino acids + 3 stop codons)
Universal	The same codons code for the same amino acids in almost all organisms (supports common ancestry)
Non-overlapping	Each base is part of only one codon; codons are read sequentially
Triplet	Each codon consists of 3 nucleotides

Start codon: AUG (methionine). Stop codons: UAA, UAG, UGA (do not code for any amino acid; signal termination of translation).

22. Water: Properties and Biological Significance

22.1 Key Properties of Water

Property	Cause	Biological Importance
High specific heat capacity	Many hydrogen bonds must be broken to raise temperature	Water temperature changes slowly; provides a stable environment for aquatic organisms; buffers temperature fluctuations in organisms
High latent heat of vaporisation	Many hydrogen bonds must be broken for water to evaporate	Effective cooling mechanism (sweating, panting)
Cohesion and surface tension	Hydrogen bonds between water molecules	Water column can be pulled up xylem vessels (cohesion-tension theory); surface tension supports some organisms (e.g., pond skaters)
High solvent power	Polar nature; water molecules form hydration shells around ions and polar molecules	Biochemical reactions occur in aqueous solution; transport of dissolved substances (sugars, amino acids, ions, gases) in blood and sap
Density anomaly (ice floats)	Hydrogen bonds form a regular lattice in ice, creating a more open structure than liquid water	Ice insulates the water below, preventing complete freezing of aquatic habitats in winter
Transparency	Water absorbs light weakly in the visible spectrum	Light can penetrate water, allowing photosynthesis in aquatic plants and algae
High surface tension	Cohesion between water molecules at the surface	Enables some organisms to walk on water; draws water up capillary tubes (e.g., in soil)

22.2 Water Potential

Water potential ($\Psi$) is the tendency of water to move from one area to another. Pure water at standard temperature and pressure has a water potential of 0.

$\Psi = \Psi_s + \Psi_p$

Where $\Psi_s$ = solute potential (always negative or zero; solutes lower water potential) and $\Psi_p$ = pressure potential (positive in plant cells due to cell wall pressure; negative in xylem during transpiration).

Water always moves from a region of higher water potential to a region of lower water potential.

22.3 Inorganic Ions in Biology

Ion	Role	Deficiency Consequence
$\mathrm{Na^+}$	Nerve impulse transmission; $\mathrm{Na^+/K^+}$ pump; co-transport of glucose/amino acids	Hyponatraemia (confusion, seizures)
$\mathrm{K^+}$	Nerve impulse transmission; stomatal opening; enzyme cofactor	Hypokalaemia (muscle weakness, cardiac arrhythmias)
$\mathrm{Ca^{2+}}$	Bone and teeth formation; blood clotting (factor IV); muscle contraction; second messenger	Osteoporosis; tetany
$\mathrm{Mg^{2+}}$	Chlorophyll component; enzyme cofactor (e.g., Rubisco, DNA polymerases)	Chlorosis (yellowing of leaves); muscle cramps
$\mathrm{Fe^{2+/3+}}$	Haemoglobin component (oxygen transport); cytochromes (ETC)	Anaemia (iron deficiency); chlorosis
$\mathrm{PO_4^{3-}}$	ATP, DNA, RNA, phospholipids; bone and teeth	Rickets (children); osteomalacia (adults); poor root development in plants
$\mathrm{NO_3^-}$	Amino acid and protein synthesis; nucleotide synthesis	Chlorosis (nitrogen deficiency); poor growth
$\mathrm{I^-}$	Thyroid hormone synthesis ($\mathrm{T_3}$, $\mathrm{T_4}$)	Goitre; cretinism (in infants); hypothyroidism

23. Biochemical Tests

23.1 Tests for Biological Molecules

Test	Substance Detected	Procedure	Positive Result
Benedict's test	Reducing sugars (e.g., glucose, maltose)	Add Benedict's reagent (copper sulphate in alkaline solution); heat in water bath (95 degrees C, 5 min)	Brick-red precipitate (copper(I) oxide); colour change: blue $\to$ green $\to$ yellow $\to$ orange $\to$ brick red
Iodine test	Starch	Add iodine solution to sample	Blue-black colour
Biuret test	Proteins (peptide bonds)	Add Biuret reagent (NaOH + copper sulphate); mix	Colour change from blue to violet/purple
Emulsion test	Lipids	Dissolve in ethanol; pour into water	White emulsion (cloudy layer)
DCPIP test	Vitamin C (ascorbic acid)	Add DCPIP solution (blue) to sample	DCPIP decolourises (blue $\to$ colourless)

23.2 Quantitative Biochemical Tests

Glucose concentration using colorimetry:

1. Prepare a series of glucose solutions of known concentration (standards).
2. Add Benedict's reagent to each standard and to the unknown sample.
3. Heat all tubes in a water bath.
4. Filter each tube to collect the precipitate.
5. Measure the absorbance of each filtrate using a colorimeter (red filter, ~620 nm).
6. Plot absorbance vs concentration for the standards (calibration curve).
7. Read the concentration of the unknown from the calibration curve.

23.3 Chromatography

Paper chromatography of photosynthetic pigments:

1. Extract pigments from leaves by grinding in solvent (e.g., propanone).
2. Spot the extract on a pencil line near the bottom of chromatography paper.
3. Place the paper in a beaker with solvent (e.g., petroleum ether:acetone, 9:1). The solvent level must be below the spot.
4. Allow the solvent to rise by capillary action.
5. Remove the paper when the solvent front is near the top.
6. Mark the solvent front and the position of each pigment spot.
7. Calculate the $R_f$ value for each pigment:

$R_f = \frac◆LB◆\text{distance travelled by pigment}◆RB◆◆LB◆\text{distance travelled by solvent front}◆RB◆$

Typical $R_f$ values (petroleum ether:acetone solvent):

Pigment	Colour	Approximate $R_f$
Carotene	Yellow-orange	0.95 (most non-polar; travels furthest)
Xanthophyll	Yellow	0.71
Chlorophyll $a$	Blue-green	0.59
Chlorophyll $b$	Yellow-green	0.42 (most polar; travels least)

24. Lipids: Structure and Function

24.1 Triglycerides

Triglycerides (triacylglycerols) consist of one glycerol molecule esterified to three fatty acid molecules:

$\text{Glycerol} + 3\text{ fatty acids} \rightleftharpoons \text{triglyceride} + 3\text{H}_2\text{O}$

Fatty acid types:

Type	Bonds	Melting Point	State at Room Temperature	Examples
Saturated	No C=C double bonds	Higher	Solid	Butter, lard, palm oil
Monounsaturated	One C=C double bond	Lower	Liquid (oil)	Olive oil, rapeseed oil
Polyunsaturated	Multiple C=C double bonds	Lowest	Liquid	Fish oil, sunflower oil, walnut oil

Cis double bonds cause kinks in the fatty acid chain, preventing close packing and lowering the melting point. Trans fats (hydrogenated vegetable oils) have trans double bonds (no kink), so they pack more tightly and behave like saturated fats. Trans fats increase LDL ("bad") cholesterol and decrease HDL ("good") cholesterol, increasing the risk of CHD.

24.2 Phospholipids

Phospholipids are modified triglycerides: one fatty acid is replaced by a phosphate group (which may be bonded to additional groups such as choline, serine, or inositol).

• Head region: hydrophilic (water-loving) -- phosphate group and any attached groups.
• Tail region: hydrophobic (water-fearing) -- two fatty acid chains.

This amphipathic nature makes phospholipids ideal for forming cell membranes: the hydrophilic heads face the aqueous environment (inside and outside the cell), while the hydrophobic tails face each other in the interior of the bilayer.

24.3 Cholesterol

Cholesterol is a steroid lipid found in animal cell membranes:

• Located between phospholipid molecules in the bilayer.
• Hydrophobic ring structure interacts with the hydrophobic tails of phospholipids.
• Hydrophilic -OH group interacts with the hydrophilic phosphate heads.
• Functions in the membrane: regulates membrane fluidity (acts as a "fluidity buffer" -- prevents crystallisation at low temperatures and limits excessive fluidity at high temperatures); reduces permeability to water-soluble molecules and ions.
• Other roles: precursor for steroid hormones (testosterone, oestrogen, cortisol); precursor for bile salts (emulsify fats in the small intestine); precursor for vitamin D (synthesised in the skin by UV light).

24.4 Lipid Digestion and Absorption

1. Emulsification: bile salts (synthesised in the liver from cholesterol, stored in the gall bladder) emulsify lipids in the small intestine, breaking large fat droplets into smaller droplets (micelles). This increases the surface area for lipase action.
2. Lipase: pancreatic lipase hydrolyses triglycerides into monoglycerides and fatty acids.
3. Micelle formation: monoglycerides and fatty acids form micelles with bile salts, which transport them to the epithelial cells of the small intestine.
4. Absorption: monoglycerides and fatty acids diffuse across the epithelial cell membrane into the cell.
5. Re-esterification: inside the epithelial cell, monoglycerides and fatty acids are recombined into triglycerides.
6. Chylomicron formation: triglycerides are packaged with cholesterol and proteins into chylomicrons (lipoprotein particles), which enter the lacteals (lymphatic capillaries in the villi).
7. Transport: chylomicrons enter the lymphatic system and are transported to the bloodstream via the thoracic duct.

25. Carbohydrates: Structure and Function

25.1 Monosaccharides

Monosaccharides are the simplest carbohydrates (single sugar units):

Monosaccharide	Formula	Key Feature	Function
Glucose	\mathrm{C_6H_{12O_6}	Hexose; aldose (contains aldehyde group)	Main respiratory substrate; blood sugar
Fructose	\mathrm{C_6H_{12O_6}	Hexose; ketose (contains ketone group)	Found in fruits; sweetest natural sugar
Galactose	\mathrm{C_6H_{12O_6}	Hexose; aldose	Component of lactose (milk sugar)
Ribose	\mathrm{C_5H_{10O_5}	Pentose	Component of RNA, ATP, NAD$^+$
Deoxyribose	\mathrm{C_5H_{10O_4}	Pentose (one fewer O than ribose)	Component of DNA

Isomerism:

• $\alpha$ and $\beta$ forms: glucose exists in two forms depending on the position of the -OH group on carbon 1. In $\alpha$-glucose, the -OH is below the plane; in $\beta$-glucose, it is above the plane. This difference has major structural consequences: starch is made of $\alpha$-glucose (helical, digestible); cellulose is made of $\beta$-glucose (straight chains, indigestible by humans).

25.2 Disaccharides

Disaccharide	Component Monosaccharides	Bond	Found In
Maltose	Glucose + glucose	$\alpha$-1,4 glycosidic	Malt (germinating barley); digestion product of starch
Sucrose	Glucose + fructose	$\alpha$-1,2 glycosidic	Cane sugar, beet sugar; transported in phloem
Lactose	Glucose + galactose	$\beta$-1,4 glycosidic	Milk

Lactose intolerance: deficiency of lactase (the enzyme that hydrolyses lactose into glucose and galactose) causes undigested lactose to reach the large intestine, where bacteria ferment it, producing gas ($\mathrm{CO_2}$, $\mathrm{CH_4}$) and causing bloating, flatulence, and diarrhoea. Approximately 70% of the world's adult population has some degree of lactase deficiency.

25.3 Polysaccharides

Polysaccharide	Monomer	Bonds	Structure	Function
Starch (amylose)	$\alpha$-glucose	$\alpha$-1,4 glycosidic	Coiled (helical)	Energy storage in plants; compact; insoluble (does not affect water potential)
Starch (amylopectin)	$\alpha$-glucose	$\alpha$-1,4 and $\alpha$-1,6 (branches)	Branched	Energy storage; branched structure provides more ends for enzyme action
Glycogen	$\alpha$-glucose	$\alpha$-1,4 and $\alpha$-1,6 (more branched)	Highly branched	Energy storage in animals; more branches than amylopectin for rapid glucose release
Cellulose	$\beta$-glucose	$\beta$-1,4 glycosidic	Straight chains; hydrogen bonds between chains	Structural (cell walls); tensile strength; not digestible by most animals
Chitin	N-acetylglucosamine	$\beta$-1,4 glycosidic	Similar to cellulose but with amino group	Exoskeleton of arthropods; fungal cell walls

26. Enzymes: Inhibition and Allosteric Regulation

26.1 Non-Competitive Inhibition Worked Example

An enzyme has $V_{\max} = 50\ \mu\mathrm{mol\ min^{-1}}$ and $K_m = 5\ \mathrm{mM}$ without inhibitor. With a non-competitive inhibitor present, $V_{\max}$ is reduced to $25\ \mu\mathrm{mol\ min^{-1}}$ and $K_m$ remains $5\ \mathrm{mM}$.

At $[S] = 5\ \mathrm{mM}$:

Without inhibitor: $v = \frac◆LB◆50 \times 5◆RB◆◆LB◆5 + 5◆RB◆ = 25\ \mu\mathrm{mol\ min^{-1}}$ (50% of $V_{\max}$).

With inhibitor: $v = \frac◆LB◆25 \times 5◆RB◆◆LB◆5 + 5◆RB◆ = 12.5\ \mu\mathrm{mol\ min^{-1}}$ (50% of new $V_{\max}$).

The rate is halved, and increasing $[S]$ cannot overcome the inhibition.

26.2 Competitive Inhibition Worked Example

The same enzyme ($V_{\max} = 50\ \mu\mathrm{mol\ min^{-1}}$, $K_m = 5\ \mathrm{mM}$) with a competitive inhibitor. $V_{\max}$ remains 50, but apparent $K_m$ increases to $15\ \mathrm{mM}$.

At $[S] = 5\ \mathrm{mM}$:

Without inhibitor: $v = \frac◆LB◆50 \times 5◆RB◆◆LB◆5 + 5◆RB◆ = 25\ \mu\mathrm{mol\ min^{-1}}$.

With inhibitor: $v = \frac◆LB◆50 \times 5◆RB◆◆LB◆15 + 5◆RB◆ = 12.5\ \mu\mathrm{mol\ min^{-1}}$.

At $[S] = 50\ \mathrm{mM}$:

Without inhibitor: $v = \frac◆LB◆50 \times 50◆RB◆◆LB◆5 + 50◆RB◆ = 45.5\ \mu\mathrm{mol\ min^{-1}}$.

With inhibitor: $v = \frac◆LB◆50 \times 50◆RB◆◆LB◆15 + 50◆RB◆ = 38.5\ \mu\mathrm{mol\ min^{-1}}$.

At very high $[S]$, the difference between inhibited and uninhibited rates becomes negligible (both approach $V_{\max} = 50$).

26.3 Allosteric Regulation

Many enzymes are regulated by molecules that bind at sites other than the active site (allosteric sites):

Regulator	Effect	Example
Allosteric activator	Binds to allosteric site; stabilises the active conformation of the enzyme; increases substrate affinity; shifts curve to the left	Fructose-2,6-bisphosphate activates phosphofructokinase (PFK) in glycolysis
Allosteric inhibitor	Binds to allosteric site; stabilises the inactive conformation; decreases substrate affinity; shifts curve to the right	ATP inhibits PFK (feedback inhibition when energy is abundant)

Cooperativity: some allosteric enzymes have multiple subunits. Binding of a substrate molecule to one subunit induces a conformational change that increases the affinity of the remaining subunits for substrate (positive cooperativity). This produces a sigmoidal (S-shaped) substrate-velocity curve, similar to the oxygen dissociation curve of haemoglobin.

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tip

Diagnostic Test Ready to test your understanding of Biological Molecules? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Biological Molecules with other biology topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

27. Proteins: Fibrous vs Globular

27.1 Fibrous Proteins

Fibrous proteins have an elongated, rod-like structure and are structural:

Protein	Secondary Structure	Location	Function
Collagen	Triple helix (3 polypeptide chains)	Tendons, skin, bone, cartilage, blood vessel walls	Tensile strength; structural framework
Keratin	$\alpha$-helix (coiled-coil of two $\alpha$-helices)	Hair, nails, skin (outer layer), feathers, hooves, horns	Protective; mechanical strength; waterproofing
Elastin	Cross-linked, random coil	Arterial walls, lungs, skin	Elasticity (ability to stretch and recoil)
Fibroin	$\beta$-pleated sheet	Spider silk, silkworm silk	High tensile strength; lightweight; very strong for its mass

27.2 Globular Proteins

Globular proteins have a compact, roughly spherical 3D structure and are functional:

Protein	Function	Structure
Haemoglobin	$\mathrm{O_2}$ transport in blood	Quaternary ($\alpha_2\beta_2$); globin fold in each subunit
Insulin	Hormone (regulates blood glucose)	Two polypeptide chains (A and B) joined by disulphide bonds
Enzymes	Catalyse biochemical reactions	Active site with specific 3D conformation; induced fit model
Antibodies	Immune defence ( recognise antigens)	Y-shaped; variable region (antigen binding) and constant region (effector function)
Myoglobin	$\mathrm{O_2}$ storage in muscle	Single polypeptide with a haem group; higher $\mathrm{O_2}$ affinity than haemoglobin
Lysozyme	Enzyme that breaks down bacterial cell walls (peptidoglycan)	Compact globular protein with a cleft (active site)

28. Enzyme Kinetics: Michaelis-Menten in Detail

28.1 The Michaelis-Menten Equation

$v = \frac◆LB◆V_{max} \times [S]◆RB◆◆LB◆K_m + [S]◆RB◆$

Symbol	Meaning
$v$	Initial rate of reaction
$V_{max}$	Maximum rate of reaction (when all active sites are saturated)
$[S]$	Substrate concentration
$K_m$	Substrate concentration at which $v = \frac{V_{max}}{2}$; a measure of the enzyme's affinity for the substrate

28.2 Interpreting $K_m$

$K_m$ Value	Enzyme-Substrate Affinity	Example
Low $K_m$	High affinity (enzyme reaches $\frac{V_{max}}{2}$ at low $[S]$)	Hexokinase (glucose phosphorylation; $K_m \approx 0.1\ \mathrm{mM}$)
High $K_m$	Low affinity (enzyme needs high $[S]$ to reach $\frac{V_{max}}{2}$)	Glucokinase (liver glucose phosphorylation; $K_m \approx 10\ \mathrm{mM}$; ensures liver only processes glucose when blood glucose is high)

28.3 Lineweaver-Burk Plot

Taking the reciprocal of the Michaelis-Menten equation:

$\frac{1}{v} = \frac{K_m}{V_{max}} \cdot \frac{1}{[S]} + \frac{1}{V_{max}}$

This is a straight line: $y = mx + c$ where:

• $y$-intercept $= \frac{1}{V_{max}}$
• $x$-intercept $= -\frac{1}{K_m}$
• Gradient $= \frac{K_m}{V_{max}}$

Inhibitor Type	Effect on $V_{max}$	Effect on $K_m$	Lineweaver-Burk Plot
Competitive	No change (at very high $[S]$, $V_{max}$ is reached)	Increases	Intersects on y-axis; different slopes
Non-competitive	Decreases (maximum rate is reduced)	No change	Intersects on x-axis; same slope

29. Lipids: Structure and Function in Detail

29.1 Triglyceride Structure

A triglyceride consists of one glycerol molecule esterified to three fatty acid chains:

$\text{Glycerol} + 3\ \text{Fatty acids} \to \text{Triglyceride} + 3\ \mathrm{H_2O}$

Fatty Acid Type	Bond Type	State at Room Temperature	Examples
Saturated	Only C--C single bonds; no double bonds	Solid	Butter, lard, palm oil
Monounsaturated	One C=C double bond	Liquid (but can solidify when cooled)	Olive oil, rapeseed oil
Polyunsaturated	Two or more C=C double bonds	Liquid	Sunflower oil, fish oil, flaxseed oil

29.2 Phospholipid Structure

Phospholipids are similar to triglycerides but with one fatty acid replaced by a phosphate group:

Component	Property
Glycerol	Backbone (3-carbon alcohol)
Fatty acid 1 (usually saturated)	Hydrophobic tail
Fatty acid 2 (usually unsaturated)	Hydrophobic tail; kink from double bond increases membrane fluidity
Phosphate group	Hydrophilic head (negatively charged at physiological pH)
Additional group (e.g., choline, serine, ethanolamine)	Attached to phosphate; varies between phospholipid types

29.3 Cholesterol Structure and Role

Feature	Description
Structure	Steroid ring system (4 fused carbon rings) with a hydroxyl group ($-\mathrm{OH}$) and a hydrocarbon tail
Amphipathic	Small polar head ($-\mathrm{OH}$) and large non-polar body (steroid rings + tail)
Location in membrane	Interspersed between phospholipids in the bilayer; the $-\mathrm{OH}$ interacts with phosphate heads, the steroid rings with fatty acid tails

30. Water: Properties and Biological Significance

30.1 Properties of Water

Property	Cause	Biological Importance
High specific heat capacity	Many hydrogen bonds must be broken to raise temperature	Acts as a temperature buffer; organisms maintain stable internal temperatures; aquatic environments are thermally stable
High latent heat of vaporisation	Many hydrogen bonds must be broken to evaporate	Effective cooling mechanism (sweating, panting, transpiration)
High surface tension	Cohesion between water molecules at the air-water interface	Supports small organisms on water surface (pond skaters); allows water to be drawn up xylem vessels (cohesion-tension)
Cohesion	Hydrogen bonds between water molecules	Water columns in xylem vessels resist breaking (transpiration pull); water flows as a continuous stream
Adhesion	Hydrogen bonds between water and hydrophilic surfaces	Water adheres to cell walls (capillary action); helps water move through soil
Solvent properties	Polar nature of water; hydrogen bonding with solutes	Dissolves ions, sugars, amino acids, gases; enables metabolic reactions in aqueous cytoplasm and body fluids
High density (maximum at 4$\degree\mathrm{C}$)	Hydrogen bonding creates an open lattice when frozen	Ice floats on water (insulates water below; prevents lakes from freezing solid; aquatic organisms survive winter)
Transparency	Water absorbs little visible light	Light penetrates water; enables photosynthesis in aquatic ecosystems

30.2 Water as a Reactant

Process	Role of Water
Hydrolysis	Water breaks bonds (e.g., digestion of proteins, carbohydrates, lipids; ATP hydrolysis)
Photosynthesis (light-dependent)	Water is split by photolysis: $\mathrm{2H_2O \to 4H^+ + 4e^- + O_2}$ (provides electrons and protons for the ETC)
Condensation reactions	Water is a product (e.g., formation of peptide bonds, glycosidic bonds, ester bonds)

31. Biochemical Tests for Biological Molecules

31.1 Summary Table

Molecule	Test	Reagent(s)	Positive Result	Procedure
Reducing sugars	Benedict's test	Benedict's reagent (blue, contains $\mathrm{CuSO_4}$)	Brick-red precipitate ($\mathrm{Cu_2O}$)	Add 2 cm3 of food sample + 2 cm3 of Benedict's reagent; heat in boiling water bath for 5 min
Non-reducing sugars	Benedict's test (after hydrolysis)	Benedict's reagent + dilute HCl + $\mathrm{NaHCO_3}$	Brick-red precipitate (after hydrolysis)	First test with Benedict's (negative); then hydrolyse with dilute HCl (heat); neutralise with $\mathrm{NaHCO_3}$; re-test with Benedict's
Starch	Iodine test	Iodine solution (potassium iodide-iodine, $\mathrm{I_2/KI}$)	Blue-black colour	Add a few drops of iodine solution to the sample
Protein (peptide bonds)	Biuret test	Biuret reagent (NaOH + $\mathrm{CuSO_4}$)	Violet/purple colour	Add equal volumes of Biuret reagent to sample; gentle mixing; colour change within 5 min
Lipids	Emulsion test	Ethanol; water	Milky white emulsion layer	Dissolve sample in ethanol; pour into water; if lipid is present, a cloudy white emulsion forms
Vitamin C	DCPIP test	DCPIP solution (blue)	DCPIP decolourises (turns colourless)	Add DCPIP dropwise to the sample; count drops until decolourised

31.2 Quantitative Benedict's Test

A quantitative Benedict's test can estimate the concentration of reducing sugar:

Benedict's Colour	Approximate Reducing Sugar Concentration
Blue (no change)	0%
Green	0.1--0.5%
Yellow	0.5--1.0%
Orange	1.0--1.5%
Brick-red	> 1.5%

32. Carbohydrate Chemistry: Monosaccharides

32.1 Common Monosaccharides

Monosaccharide	Formula	Key Features
Glucose	$\mathrm{C_6H_{12}O_6}$	Main respiratory substrate; most common hexose; exists as $\alpha$-glucose and $\beta$-glucose (anomers; differ at C1)
Fructose	$\mathrm{C_6H_{12}O_6}$	Ketose (carbonyl group at C2); sweetest natural sugar; found in fruit
Galactose	$\mathrm{C_6H_{12}O_6}$	Component of lactose (milk sugar); differs from glucose at C4 (epimer)
Ribose	$\mathrm{C_5H_{10}O_5}$	Pentose sugar; component of RNA, ATP, NAD, FAD
Deoxyribose	$\mathrm{C_5H_{10}O_4}$	Pentose sugar; component of DNA (no hydroxyl group at C2)

32.2 $\alpha$-Glucose vs $\beta$-Glucose

Feature	$\alpha$-Glucose	$\beta$-Glucose
OH group at C1	Below the plane of the ring	Above the plane of the ring
Polymers formed	Starch (amylose + amylopectin); glycogen	Cellulose
Glycosidic bonds	$\alpha$-1,4 and $\alpha$-1,6 glycosidic bonds	$\beta$-1,4 glycosidic bonds
Digestible by humans	Yes (amylase breaks $\alpha$-glycosidic bonds)	No (humans lack cellulase; cellulose is dietary fibre)
Structure of polymer	Helical (coiled)	Straight chains; hydrogen bonding between adjacent chains

33. Nucleic Acid Structure in Detail

33.1 DNA vs RNA Comparison

Feature	DNA	RNA
Sugar	Deoxyribose (no -OH at C2)	Ribose (-OH at C2)
Bases	Adenine, thymine, cytosine, guanine	Adenine, uracil, cytosine, guanine
Strands	Double-stranded (usually)	Single-stranded (usually)
Structure	Double helix (antiparallel strands)	Variable (single strand; can fold into complex 3D shapes; some are double-stranded)
Stability	More stable (deoxyribose is less reactive; double helix protects bases)	Less stable (ribose is more reactive; single strand exposes bases to nucleases)
Location	Nucleus (and mitochondria/chloroplasts)	Nucleus (pre-mRNA); cytoplasm (mRNA, tRNA, rRNA)
Function	Genetic information storage; heredity	Protein synthesis (mRNA, tRNA, rRNA); gene regulation (miRNA, siRNA); catalytic (ribozymes)
Length	Very long (millions of base pairs)	Shorter (mRNA ~500--10,000 nucleotides; tRNA ~75 nucleotides)

33.2 Base Pairing Rules

Base Pair	Number of Hydrogen Bonds
A--T (DNA) / A--U (RNA)	2 hydrogen bonds
G--C	3 hydrogen bonds

DNA with a higher GC content is more stable (more hydrogen bonds hold the strands together; higher melting temperature).

34. Starch, Glycogen, and Cellulose

34.1 Starch

Feature	Description
Components	Two polymers: amylose and amylopectin
Amylose	Long, unbranched chain of $\alpha$-glucose; $\alpha$-1,4 glycosidic bonds; helical structure; ~20--30% of starch
Amylopectin	Branched chain of $\alpha$-glucose; $\alpha$-1,4 glycosidic bonds + $\alpha$-1,6 glycosidic bonds at branch points (every 24--30 residues); ~70--80% of starch
Function	Energy storage in plants; stored in chloroplasts (temporary) and amyloplasts (permanent storage in seeds, tubers)
Test	Iodine test: blue-black colour

34.2 Glycogen

Feature	Description
Structure	Similar to amylopectin but more highly branched ($\alpha$-1,6 branches every 8--12 residues)
Function	Energy storage in animals; stored in liver and muscle cells
Why more branched than starch	More branch points = more terminal glucose residues = faster hydrolysis by enzymes = faster glucose release (animals need rapid energy mobilisation)
Test	Iodine test: red-brown colour (different from starch)

34.3 Cellulose

Feature	Description
Structure	Long, straight, unbranched chains of $\beta$-glucose; $\beta$-1,4 glycosidic bonds; chains linked by hydrogen bonds to form microfibrils
Function	Structural support in plant cell walls
Why it is strong	$\beta$-glucose alternates orientation (180$\degree$ flip); straight chains align parallel; many hydrogen bonds between adjacent chains; microfibrils are arranged in a mesh (cross-linked with hemicellulose and lignin)
Why humans cannot digest it	No enzyme to break $\beta$-1,4 glycosidic bonds; cellulase is produced by some bacteria, fungi, and herbivores (in their gut microbiome)

35. Protein Synthesis: Translation in Detail

35.1 The Ribosome

Component	Description
Large subunit (60S in eukaryotes)	Contains three tRNA binding sites (A site, P site, E site); catalytic site for peptide bond formation
Small subunit (40S in eukaryotes)	Binds to the mRNA; ensures the mRNA is in the correct reading frame
A site (aminoacyl site)	Incoming aminoacyl-tRNA binds here
P site (peptidyl site)	tRNA carrying the growing polypeptide chain is held here
E site (exit site)	Deacylated tRNA exits the ribosome here

35.2 Steps in Translation

Step	What Happens
1. Initiation	Small ribosomal subunit binds to the mRNA at the 5' cap; scans along the mRNA to find the start codon (AUG); initiator tRNA (carrying methionine) binds to the start codon in the P site; large ribosomal subunit joins
2. Elongation	Aminoacyl-tRNA enters the A site (matching the next codon); peptide bond forms between the amino acid in the P site and the amino acid in the A site (catalysed by peptidyl transferase); ribosome translocates by one codon: tRNA in P site moves to E site (exits); tRNA in A site moves to P site
3. Termination	When a stop codon (UAA, UAG, UGA) enters the A site, no tRNA matches it; release factor protein binds; polypeptide is released; ribosome subunits dissociate

35.3 One Gene, One Polypeptide

Feature	Description
Amino acids	20 different amino acids used in protein synthesis
Codons	64 possible codons (4 nucleotides to the power of 3); 61 code for amino acids; 3 are stop codons
Degeneracy	Most amino acids are coded for by more than one codon (e.g., leucine: 6 codons; serine: 4 codons)
Universality	The genetic code is nearly universal (same codons code for the same amino acids in almost all organisms); evidence for common ancestry

36. Enzyme Inhibition: Detailed Calculations

36.1 Calculating the Effect of a Competitive Inhibitor

Scenario: An enzyme has $K_m = 5\ \mathrm{mM}$ and $V_{max} = 100\ \mu\mathrm{mol/min}$. A competitive inhibitor is added that doubles the apparent $K_m$.

Parameter	Without Inhibitor	With Inhibitor
$K_m$	5 mM	10 mM (doubled)
$V_{max}$	100 $\mu$mol/min	100 $\mu$mol/min (unchanged)
$v$ at $[S] = 5\ \mathrm{mM}$	$\frac◆LB◆100 \times 5◆RB◆◆LB◆5 + 5◆RB◆ = 50\ \mu\mathrm{mol/min}$	$\frac◆LB◆100 \times 5◆RB◆◆LB◆10 + 5◆RB◆ = 33.3\ \mu\mathrm{mol/min}$
$v$ at $[S] = 50\ \mathrm{mM}$	$\frac◆LB◆100 \times 50◆RB◆◆LB◆5 + 50◆RB◆ = 90.9\ \mu\mathrm{mol/min}$	$\frac◆LB◆100 \times 50◆RB◆◆LB◆10 + 50◆RB◆ = 83.3\ \mu\mathrm{mol/min}$

At very high $[S]$, both approach $V_{max}$.

36.2 Calculating the Effect of a Non-Competitive Inhibitor

Scenario: A non-competitive inhibitor reduces $V_{max}$ to 60% of its original value.

Parameter	Without Inhibitor	With Inhibitor
$K_m$	5 mM	5 mM (unchanged)
$V_{max}$	100 $\mu$mol/min	60 $\mu$mol/min (reduced)
$v$ at $[S] = 5\ \mathrm{mM}$	50 $\mu$mol/min	$\frac◆LB◆60 \times 5◆RB◆◆LB◆5 + 5◆RB◆ = 30\ \mu\mathrm{mol/min}$
$v$ at $[S] = 50\ \mathrm{mM}$	90.9 $\mu$mol/min	$\frac◆LB◆60 \times 50◆RB◆◆LB◆5 + 50◆RB◆ = 54.5\ \mu\mathrm{mol/min}$

Even at very high $[S]$, the inhibited enzyme never reaches the original $V_{max}$.

37. Lipids: Triglyceride Synthesis and Breakdown

37.1 Triglyceride Synthesis (Esterification)

$\text{Glycerol} + 3\ \text{Fatty acids} \xrightarrow{\text{esterase}} \text{Triglyceride} + 3\ \mathrm{H_2O}$

This is a condensation reaction (water is released).

37.2 Triglyceride Breakdown (Hydrolysis)

$\text{Triglyceride} + 3\ \mathrm{H_2O} \xrightarrow{\text{lipase}} \text{Glycerol} + 3\ \text{Fatty acids}$

This is a hydrolysis reaction (water is used to break bonds).

37.3 Role of Bile Salts in Fat Digestion

Feature	Description
Produced by	Liver; stored in the gall bladder; released into the small intestine
Function	Emulsify lipids (break large fat droplets into smaller droplets); increases surface area for lipase enzymes
Mechanism	Bile salts are amphipathic (have both hydrophilic and hydrophobic regions); they coat fat droplets, preventing them from coalescing
Not an enzyme	Bile salts do not chemically break down lipids; they only physically emulsify them; the actual chemical breakdown is done by lipase
Recycled	Bile salts are reabsorbed in the ileum and returned to the liver (enterohepatic circulation)

38. Amino Acid Structure and Properties

38.1 General Structure

All amino acids share the same basic structure:

• An amino group ($-\mathrm{NH_2}$)
• A carboxyl group ($-\mathrm{COOH}$)
• A hydrogen atom ($-\mathrm{H}$)
• An R group (side chain) -- this is different for each of the 20 amino acids

38.2 Properties of the R Group

Property	R Groups That Have It	Examples
Hydrophobic (non-polar)	Non-polar side chains (alkyl groups, aromatic rings)	Glycine, alanine, valine, leucine, isoleucine, phenylalanine, tryptophan, methionine, proline
Hydrophilic (polar)	Polar side chains (can form hydrogen bonds)	Serine, threonine, cysteine, tyrosine, asparagine, glutamine
Charged (acidic)	Negatively charged at physiological pH	Aspartic acid, glutamic acid
Charged (basic)	Positively charged at physiological pH	Lysine, arginine, histidine
Aromatic	Contain a benzene ring	Phenylalanine, tyrosine, tryptophan
Sulphur-containing	Contain sulphur	Cysteine (forms disulphide bonds), methionine

38.3 Peptide Bond Formation

Amino acids join by condensation reactions (peptide bond formation):

$\text{Amino acid 1} + \text{Amino acid 2} \to \text{Dipeptide} + \mathrm{H_2O}$

The peptide bond forms between the carboxyl group of one amino acid and the amino group of the next. The bond is a C--N bond (specifically, a carbon-nitrogen single bond with partial double-bond character due to resonance).

39. Lipids in Detail

39.1 Triglyceride Synthesis and Breakdown

Process	Type of Reaction	Enzyme	Products
Synthesis	Condensation	Lipase (in reverse direction)	Triglyceride + 3 $\mathrm{H_2O}$
Hydrolysis	Hydrolysis	Lipase	Glycerol + 3 fatty acids
Digestion	Hydrolysis (in small intestine)	Pancreatic lipase	Monoglycerides + fatty acids (which are then absorbed)

39.2 Phospholipids and Membrane Properties

Property	Effect on Membrane
Amphipathic (hydrophilic head, hydrophobic tails)	Spontaneously forms bilayers in aqueous environments
Saturated fatty acid tails	Straight chains; pack tightly; more rigid membrane; higher melting point
Unsaturated fatty acid tails	Kinked chains (cis double bonds); prevent tight packing; more fluid membrane; lower melting point
Cholesterol	Fits between phospholipid tails; acts as a fluidity buffer: decreases fluidity at high temperatures, prevents rigidity at low temperatures

39.3 Comparison of Lipid Types

Lipid Type	Elements	Structure	Function
Triglyceride	C, H, O	Glycerol + 3 fatty acids	Energy storage; insulation; buoyancy
Phospholipid	C, H, O, P (and N)	Glycerol + 2 fatty acids + phosphate group	Cell membrane structure
Cholesterol	C, H, O	Four fused carbon rings + hydroxyl group	Membrane fluidity; precursor for steroid hormones and bile
Waxes	C, H, O	Long-chain fatty acid + long-chain alcohol	Waterproof coating on leaves (cuticle) and insect exoskeletons

40. Carbohydrate Digestion

40.1 Enzymes of Carbohydrate Digestion

Enzyme	Site of Action	Substrate	Products
Salivary amylase	Mouth	Starch	Maltose (and some dextrins)
Pancreatic amylase	Small intestine (duodenum)	Starch	Maltose
Maltase	Brush border of small intestine	Maltose	Glucose
Sucrase	Brush border of small intestine	Sucrose	Glucose + fructose
Lactase	Brush border of small intestine	Lactose	Glucose + galactose

40.2 Lactose Intolerance

Feature	Description
Cause	Deficiency of lactase enzyme in the small intestine brush border
Effect	Lactose passes undigested to the large intestine; bacteria ferment lactose; produces gas ($\mathrm{CO_2}$, $\mathrm{CH_4}$, $\mathrm{H_2}$) and organic acids
Symptoms	Bloating, flatulence, abdominal pain, diarrhoea
Prevalence	~70% of the world's population lose lactase activity after weaning; highest prevalence in East Asian, West African, and Native American populations
Genetics	Lactase persistence (ability to digest lactose as an adult) is caused by mutations in the regulatory region of the LCT gene; evolved independently in European, African, and Middle Eastern populations (convergent evolution)

41. Protein Structure in Detail

41.1 Primary Structure

Feature	Description
What it is	The sequence of amino acids in a polypeptide chain
Bonds involved	Peptide bonds (covalent) between adjacent amino acids
Importance	Determines all higher levels of structure; a change in one amino acid (mutation) can alter the entire protein's shape and function

41.2 Secondary Structure

Type	Description
Alpha helix	Polypeptide chain coils into a spiral; hydrogen bonds form between the $\mathrm{C=O}$ of one amino acid and the $\mathrm{N-H}$ of another amino acid four residues later; these H-bonds run parallel to the helix axis
Beta pleated sheet	Polypeptide chains lie alongside each other; hydrogen bonds form between $\mathrm{C=O}$ and $\mathrm{N-H}$ groups of adjacent chains; the sheet can be parallel or antiparallel

41.3 Tertiary Structure

Feature	Description
What it is	The overall 3D shape of a single polypeptide chain; formed by folding of the secondary structure
Bonds involved	Disulfide bridges (strong covalent bonds between cysteine residues); ionic bonds (between charged R groups); hydrogen bonds (between polar R groups); hydrophobic interactions (non-polar R groups cluster in the centre of the protein, away from water)
Importance	Determines the protein's function; the active site of an enzyme is part of the tertiary structure

41.4 Quaternary Structure

Feature	Description
What it is	The arrangement of two or more polypeptide chains (subunits) into a single functional protein
Bonds involved	Same as tertiary structure (H-bonds, ionic bonds, hydrophobic interactions, disulfide bridges) between the different subunits
Examples	Haemoglobin (4 subunits: 2 alpha, 2 beta); collagen (3 polypeptide chains wound together)
Not all proteins	Have quaternary structure; many functional proteins consist of a single polypeptide chain (e.g., myoglobin, lysozyme)

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