Molecular Biology

1. Water

Molecular Structure and Properties

Water ($\mathrm{H}_2\mathrm{O}$) is a polar molecule: the oxygen atom carries a partial negative charge ($\delta-$) and each hydrogen carries a partial positive charge ($\delta+$). The bond angle is approximately $104.5^\circ$.

The polarity gives rise to hydrogen bonding between water molecules: the $\delta+$ hydrogen of one molecule is attracted to the $\delta-$ oxygen of another. Each water molecule can form up to four hydrogen bonds.

Properties Arising from Hydrogen Bonding

Property	Significance
Cohesion	Water molecules stick together; enables surface tension and transport of water in xylem (transpiration pull).
Adhesion	Water sticks to other polar surfaces (cell walls); aids capillary action.
High specific heat capacity ($4.18\;\mathrm{J\;g^{-1}\;^\circ C^{-1}}$)	Water resists temperature change; buffers organisms against thermal fluctuations.
High latent heat of vaporisation ($2260\;\mathrm{J\;g^{-1}}$)	Evaporation of sweat provides effective cooling.
High latent heat of fusion	Ice formation releases heat, moderating temperature in aquatic habitats.
Maximum density at $4^\circ\mathrm{C}$	Ice floats, insulating water below and allowing aquatic life to survive winter.
Excellent solvent	Polar and ionic substances dissolve; enables metabolic reactions in cytoplasm and transport in blood.
Transparency	Light penetrates water, allowing photosynthesis in aquatic ecosystems.

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2. Carbohydrates

General Formula

Carbohydrates have the general formula $\mathrm{C}_x(\mathrm{H}_2\mathrm{O})_y$. They contain C, H, and O in the ratio $1:2:1$.

Monosaccharides

The simplest carbohydrates; single sugar units. Examples: glucose ($\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$), fructose, galactose, ribose ($\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}_5$).

• Glucose: exists as alpha ($\alpha$)-glucose (OH on C1 below the plane) and beta ($\beta$)-glucose (OH on C1 above the plane). This structural difference leads to fundamentally different polymers.
• Ribose: a pentose sugar; component of RNA and ATP.
• Glucose is the primary substrate for cellular respiration (glycolysis).

Disaccharides

Formed by a condensation reaction (dehydration synthesis): two monosaccharides join with the release of $\mathrm{H}_2\mathrm{O}$. Broken by hydrolysis.

Disaccharide	Monosaccharide components	Bond type
Maltose	$\alpha$-glucose + $\alpha$-glucose	$\alpha$-1,4-glycosidic
Sucrose	$\alpha$-glucose + fructose	$\alpha$-1,2-glycosidic
Lactose	Galactose + glucose	$\beta$-1,4-glycosidic

Polysaccharides

Long chains of monosaccharides linked by glycosidic bonds.

Polysaccharide	Monomer	Bond	Structure	Function
Starch	$\alpha$-glucose	$\alpha$-1,4 (amylose); $\alpha$-1,4 and $\alpha$-1,6 (amylopectin)	Helical (amylose), branched (amylopectin)	Energy storage in plants; compact, insoluble
Glycogen	$\alpha$-glucose	$\alpha$-1,4 and $\alpha$-1,6	Highly branched	Energy storage in animals; rapid glucose release
Cellulose	$\beta$-glucose	$\beta$-1,4	Straight, unbranched chains linked by H-bonds forming microfibrils	Structural support in plant cell walls

Key distinction: $\alpha$-glucose polymers are coiled and used for energy storage; $\beta$-glucose polymers are straight and form tough structural fibres.

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3. Lipids

Triglycerides

Formed from one glycerol molecule (3-carbon alcohol) and three fatty acid molecules, joined by ester bonds via condensation reactions.

$$\mathrm{Glycerol} + 3\;\mathrm{Fatty Acids} \xrightarrow{\mathrm{condensation}} \mathrm{Triglyceride} + 3\;\mathrm{H}_2\mathrm{O}$$

• Saturated fatty acids: no C=C double bonds; straight chains; solid at room temperature (e.g., butter). Derived mainly from animal sources.
• Unsaturated fatty acids: one or more C=C double bonds; kinked chains; liquid at room temperature (e.g., olive oil). Derived mainly from plant sources.
• Monounsaturated: one C=C bond. Polyunsaturated: two or more C=C bonds.

Functions: long-term energy storage (more energy per gram than carbohydrates: $\approx 39\;\mathrm{kJ\;g^{-1}}$ vs $\approx 17\;\mathrm{kJ\;g^{-1}}$), insulation, buoyancy, and protection of organs.

Phospholipids

Similar to triglycerides but one fatty acid is replaced by a phosphate group. This gives phospholipids an amphipathic nature: the phosphate head is hydrophilic, and the two fatty acid tails are hydrophobic. Phospholipids form the bilayer of all cell membranes.

Cholesterol

A steroid lipid that fits between phospholipids in animal cell membranes, modulating fluidity. Cholesterol is also a precursor for steroid hormones (testosterone, oestrogen, cortisol) and bile salts.

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4. Proteins

Amino Acids

Proteins are polymers of amino acids. There are 20 standard amino acids, each with a central alpha-carbon bonded to:

• An amino group ($\mathrm{NH}_2$)
• A carboxyl group ($\mathrm{COOH}$)
• A hydrogen atom
• A variable R group (side chain) that determines the amino acid's properties

Peptide bonds form between the carboxyl group of one amino acid and the amino group of another via a condensation reaction. Hydrolysis breaks peptide bonds.

Protein Structure

Level	Description
Primary	Linear sequence of amino acids, determined by the gene. The primary structure dictates all higher levels.
Secondary	Local folding patterns stabilised by hydrogen bonds between the peptide backbone. Common patterns: alpha-helix (coiled) and beta-pleated sheet.
Tertiary	Overall 3D shape of a single polypeptide, stabilised by: hydrogen bonds, ionic (electrostatic) bonds, disulfide bridges (between cysteine residues), hydrophobic interactions, and van der Waals forces.
Quaternary	Assembly of two or more polypeptide subunits (e.g., haemoglobin has four subunits: $2\alpha + 2\beta$).

Denaturation

Disruption of secondary, tertiary, or quaternary structure by heat, pH change, or chemical agents, causing loss of function. The primary structure (amino acid sequence) is not altered.

Functions of Proteins

• Enzymes: biological catalysts (e.g., amylase, DNA polymerase).
• Structural: collagen (connective tissue), keratin (hair, nails).
• Transport: haemoglobin ($\mathrm{O}_2$ transport), channel proteins.
• Hormones: insulin, growth hormone.
• Antibodies: immunoglobulins for immune defence.
• Movement: actin and myosin in muscle contraction.

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5. Enzymes

Nature of Enzymes

Enzymes are globular proteins that act as biological catalysts: they lower the activation energy ($E_a$) of a reaction without being consumed.

Definition. The activation energy is the minimum energy required for reactants to reach the transition state and form products.

Enzymes are specific: each enzyme catalyses a specific reaction or set of reactions due to the complementary shape of its active site to its substrate.

Lock-and-Key vs Induced Fit Model

• Lock-and-key model: the substrate fits precisely into the active site like a key in a lock. Rigid and less accurate.
• Induced fit model: the active site changes shape slightly upon substrate binding, improving the fit and facilitating catalysis. This is the currently accepted model.

Factors Affecting Enzyme Activity

Factor	Effect
Temperature	Rate increases with temperature up to an optimum (typically $35$--$40^\circ\mathrm{C}$); above this, denaturation causes a sharp decline.
pH	Each enzyme has an optimal pH; deviation alters ionisation of R groups, disrupting the active site.
Substrate concentration	Rate increases with [substrate] until all active sites are saturated ($V_{\max}$).
Enzyme concentration	Rate is proportional to [enzyme] when substrate is in excess.

Enzyme Inhibition

• Competitive inhibition: inhibitor structurally resembles the substrate and competes for the active site. Can be overcome by increasing substrate concentration. Increases the apparent $K_m$ without affecting $V_{\max}$.
• Non-competitive inhibition: inhibitor binds to an allosteric site (not the active site), changing the enzyme's shape. Cannot be overcome by increasing substrate. Decreases $V_{\max}$ without affecting $K_m$.

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6. DNA Structure

Nucleotide Structure

DNA is a polymer of nucleotides. Each nucleotide consists of:

1. A pentose sugar (deoxyribose in DNA, ribose in RNA).
2. A phosphate group.
3. A nitrogenous base: adenine (A), thymine (T), cytosine (C), or guanine (G). (RNA uses uracil, U, instead of thymine.)

The Double Helix

Watson and Crick (1953) proposed the double-helix model:

• Two antiparallel polynucleotide strands coil around a common axis.
• Chargaff's rule: A pairs with T (2 hydrogen bonds); C pairs with G (3 hydrogen bonds). $[\mathrm{A}] = [\mathrm{T}]$ and $[\mathrm{C}] = [\mathrm{G}]$.
• The sugar-phosphate backbone is on the outside; bases face inward.
• The helix is stabilised by hydrogen bonds between complementary bases and hydrophobic interactions (base stacking).

Purines (A, G) are double-ring structures; pyrimidines (T, C, U) are single-ring. A purine always pairs with a pyrimidine, maintaining a uniform helix width of $\approx 2\;\mathrm{nm}$.

DNA vs RNA

Feature	DNA	RNA
Sugar	Deoxyribose	Ribose
Bases	A, T, C, G	A, U, C, G
Strands	Double-stranded	Usually single-stranded
Structure	Double helix	Various (e.g., hairpin loops)
Location	Nucleus (mainly)	Nucleus, cytoplasm
Stability	More stable (deoxyribose lacks reactive $2'$-OH)	Less stable (ribose $2'$-OH is reactive)
Function	Genetic information storage	Protein synthesis, gene regulation

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7. DNA Replication

Semi-Conservative Replication

Meselson and Stahl (1958) demonstrated that DNA replication is semi-conservative: each new DNA molecule consists of one original ("parental") strand and one newly synthesised strand.

The Replication Process

Replication occurs during the S phase of the cell cycle. It is semi-discontinuous: the leading strand is synthesised continuously; the lagging strand is synthesised in short fragments (Okazaki fragments).

1. Helicase unwinds and separates the double helix at the origin of replication, forming a replication fork.
2. Single-strand binding proteins (SSBs) stabilise the separated strands, preventing re-annealing.
3. Topoisomerase (DNA gyrase) relieves tension ahead of the fork by cutting and rejoining DNA strands.
4. Primase synthesises a short RNA primer complementary to the template strand.
5. DNA polymerase III adds nucleotides to the $3'$ end of the primer, synthesising in the $5' \to 3'$ direction.
6. On the leading strand, synthesis is continuous toward the replication fork.
7. On the lagging strand, synthesis occurs away from the fork in Okazaki fragments ($1000$--$2000$ nucleotides in prokaryotes; $100$--$200$ in eukaryotes).
8. DNA polymerase I removes RNA primers and replaces them with DNA.
9. Ligase joins Okazaki fragments by forming phosphodiester bonds.

Fidelity

DNA polymerase has proofreading ($3' \to 5'$ exonuclease activity): if an incorrect nucleotide is added, it is removed and replaced. This gives an error rate of approximately $10^{-9}$ per base pair per replication.

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8. Transcription

Overview

Transcription is the synthesis of a complementary mRNA copy from a DNA template. It occurs in the nucleus (in eukaryotes).

Steps

1. Initiation: RNA polymerase binds to the promoter region upstream of the gene (in eukaryotes, transcription factors and the TATA box are involved).
2. Elongation: RNA polymerase unwinds the DNA and synthesises mRNA in the $5' \to 3'$ direction, using the template (antisense) strand. The coding (sense) strand has the same base sequence as the mRNA (with T replaced by U).
3. Termination: RNA polymerase detaches when it reaches a terminator sequence.

Post-Transcriptional Modification (Eukaryotes)

The primary transcript (pre-mRNA) undergoes processing:

• $5'$ capping: addition of a modified guanine nucleotide; protects mRNA and aids ribosome binding.
• $3'$ poly-A tail: addition of a polyadenine sequence; stabilises mRNA and aids export from the nucleus.
• Splicing: removal of introns (non-coding regions); exons (coding regions) are joined together. Alternative splicing allows one gene to produce multiple protein variants.

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9. Translation

Overview

Translation is the synthesis of a polypeptide chain from mRNA at the ribosome. It occurs in the cytoplasm (or on the RER for secreted proteins).

The Genetic Code

• Triplet code: each codon (three nucleotides) specifies one amino acid.
• Degenerate: most amino acids are encoded by more than one codon (e.g., leucine has six codons).
• Universal: nearly identical in all organisms (evidence for common ancestry).
• The genetic code has $64$ codons: $61$ code for amino acids; $3$ are stop codons (UAA, UAG, UGA); $1$ is the start codon (AUG, also codes for methionine).

Steps

1. Initiation: the small ribosomal subunit binds to the $5'$ end of mRNA and moves to the start codon (AUG). The initiator tRNA carrying methionine binds to the start codon. The large ribosomal subunit joins, forming the translation complex. The tRNA occupies the P site (peptidyl site).
2. Elongation:
   • A new aminoacyl-tRNA enters the A site (aminoacyl site), complementary to the next codon.
   • A peptide bond forms between the amino acid in the P site and the amino acid in the A site (catalysed by peptidyl transferase, an rRNA ribozyme).
   • The ribosome translocates by one codon: the empty tRNA moves to the E site (exit site) and is released; the tRNA carrying the growing polypeptide moves from the A site to the P site.
3. Termination: when a stop codon enters the A site, a release factor binds, causing the polypeptide to be released and the ribosome to dissociate.

Polysomes

A single mRNA can be translated simultaneously by multiple ribosomes, forming a polyribosome (polysome), increasing the rate of protein synthesis.

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Common Pitfalls

• Confusing the sense and template strands: the template strand is the one used for transcription; the sense (coding) strand has the same sequence as the mRNA (T instead of U).
• Writing DNA replication in the $3' \to 5'$ direction: DNA polymerase synthesises exclusively in the $5' \to 3'$ direction.
• Stating that "enzymes are used up in reactions": enzymes are catalysts and are regenerated at the end of each reaction cycle.
• Confusing alpha and beta glucose: the position of the OH group on C1 distinguishes them, and this difference leads to polymers with very different properties (starch vs cellulose).
• Equating introns with "junk DNA": introns can contain regulatory sequences and are involved in alternative splicing.

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Practice Problems

Question 1: Benedict's Test for Reducing Sugars

A student tests a solution containing sucrose with Benedict's reagent. The solution remains blue. The student then boils the sucrose solution with dilute hydrochloric acid, neutralises it, and repeats the test. The solution turns brick-red. Explain these observations.

Answer

Sucrose is a non-reducing disaccharide: the glycosidic bond involves both anomeric carbons, so neither monosaccharide has a free aldehyde or ketone group available to reduce $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}^+$ (Benedict's test).

Acid hydrolysis breaks the glycosidic bond, releasing glucose and fructose. Both are reducing sugars (glucose has a free aldehyde group; fructose can isomerise in solution). The reducing sugars reduce Benedict's reagent, producing a brick-red precipitate of $\mathrm{Cu}_2\mathrm{O}$.

Question 2: Enzyme Kinetics Graph

An enzyme has $V_{\max} = 40\;\mathrm{\mu mol\;min^{-1}}$ and $K_m = 5\;\mathrm{mM}$. Calculate the reaction rate when the substrate concentration is $15\;\mathrm{mM}$.

Answer

Using the Michaelis-Menten equation:

$$v = \frac{V_{\max} \cdot [S]}{K_m + [S]} = \frac{40 \times 15}{5 + 15} = \frac{600}{20} = 30\;\mathrm{\mu mol\;min^{-1}}$$

Question 3: DNA Base Composition

A sample of double-stranded DNA contains $22\%$ adenine. Determine the percentage of thymine, guanine, and cytosine.

Answer

By Chargaff's rule, $[\mathrm{A}] = [\mathrm{T}]$ and $[\mathrm{C}] = [\mathrm{G}]$.

• Thymine: $22\%$ (pairs with A)
• Adenine + Thymine total: $44\%$
• Remaining: $100\% - 44\% = 56\%$ for G and C
• Guanine: $28\%$
• Cytosine: $28\%$

Question 4: Transcription and Translation

A DNA template strand has the sequence: $3'$-TAC GGA TTA CCG-5'. Write the sequence of the mRNA produced and determine the amino acid sequence of the resulting polypeptide (use a standard genetic code table).

Answer

The mRNA is synthesised in the $5' \to 3'$ direction, complementary to the template strand:

mRNA: $5'$-AUG CCU AAU GGC-3'

Using the genetic code:

• AUG = Methionine (Met)
• CCU = Proline (Pro)
• AAU = Asparagine (Asn)
• GGC = Glycine (Gly)

Polypeptide: Met -- Pro -- Asn -- Gly

Question 5: Lipid Digestion and Absorption

Explain why triglycerides are not directly absorbed into the blood from the small intestine. Describe the role of bile salts in lipid digestion.

Answer

Triglycerides are hydrophobic and aggregate into large lipid droplets, presenting insufficient surface area for lipase action. Bile salts (produced in the liver and stored in the gallbladder) have both hydrophilic and hydrophobic regions; they emulsify fat droplets into smaller micelles, increasing the surface area available for pancreatic lipase. Lipase hydrolyses triglycerides into monoglycerides and free fatty acids, which are absorbed into intestinal epithelial cells, re-esterified into triglycerides, and packaged into chylomicrons for transport in the lymphatic system (not directly into blood).

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Worked Examples

Worked Example: Calculating DNA Quantities from Absorbance

A student measures the absorbance of a DNA solution at $260\;\mathrm{nm}$ using a spectrophotometer and obtains a reading of $0.40$. Using the relationship that an absorbance of $1.0$ corresponds to $50\;\mathrm{\mu g/mL}$ of double-stranded DNA, calculate the DNA concentration and the total mass of DNA in a $500\;\mathrm{\mu L}$ sample. Also, the absorbance at $280\;\mathrm{nm}$ is $0.22$. Calculate the $A_{260}/A_{280}$ ratio and assess the purity.

Solution

DNA concentration: $[\mathrm{DNA}] = 0.40 \times 50\;\mathrm{\mu g/mL} = 20\;\mathrm{\mu g/mL}$

Total DNA in $500\;\mathrm{\mu L}$: $\mathrm{mass} = 20\;\mathrm{\mu g/mL} \times 0.500\;\mathrm{mL} = 10\;\mathrm{\mu g}$

Purity ratio: $\frac{A_{260}}{A_{280}} = \frac{0.40}{0.22} = 1.82$

A ratio of $1.8$--$2.0$ indicates relatively pure DNA. A ratio below $1.8$ suggests protein contamination (proteins absorb strongly at $280\;\mathrm{nm}$ due to aromatic amino acids, tryptophan and tyrosine). A ratio above $2.0$ suggests RNA contamination (RNA absorbs at $260\;\mathrm{nm}$). This sample is within the acceptable range.

Worked Example: Michaelis-Menten Kinetics and Competitive Inhibition

An enzyme has $V_{\max} = 50\;\mathrm{\mu mol/min}$ and $K_m = 10\;\mathrm{mM}$ in the absence of inhibitor. A competitive inhibitor is added at a concentration that doubles the apparent $K_m$. Calculate the reaction velocity at $[S] = 10\;\mathrm{mM}$ with and without the inhibitor. At what substrate concentration does the inhibited reaction reach $90\%$ of $V_{\max}$?

Solution

Without inhibitor at $[S] = 10\;\mathrm{mM}$: $v = \frac{V_{\max} \cdot [S]}{K_m + [S]} = \frac{50 \times 10}{10 + 10} = \frac{500}{20} = 25\;\mathrm{\mu mol/min}$

With competitive inhibitor ($K_m^{\mathrm{app}} = 2 \times 10 = 20\;\mathrm{mM}$; $V_{\max}$ unchanged): $v = \frac{50 \times 10}{20 + 10} = \frac{500}{30} = 16.7\;\mathrm{\mu mol/min}$

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

Substrate concentration for $90\%$ of $V_{\max}$ with inhibitor: $0.9 \times V_{\max} = \frac{V_{\max} \cdot [S]}{K_m^{\mathrm{app}} + [S]}$ $0.9(K_m^{\mathrm{app}} + [S]) = [S]$ $0.9 K_m^{\mathrm{app}} + 0.9[S] = [S]$ $0.1[S] = 0.9 K_m^{\mathrm{app}}$ $[S] = 9 K_m^{\mathrm{app}} = 9 \times 20 = 180\;\mathrm{mM}$

At very high substrate concentrations ($[S] \gg K_m^{\mathrm{app}}$), the competitive inhibitor is outcompeted and the enzyme reaches the same $V_{\max}$ as the uninhibited reaction. This is the defining characteristic of competitive inhibition.

Worked Example: Protein Structure and Amino Acid Properties

A peptide has the sequence: Val -- Glu -- Lys -- Ala -- Cys -- Pro -- Phe. For each amino acid, state whether the side chain is polar, non-polar, acidic, or basic. Identify which amino acid(s) could form: (a) a disulfide bridge, (b) an ionic bond, (c) a hydrogen bond with water.

Solution

Amino Acid	Side Chain Property
Val	Non-polar (hydrophobic)
Glu	Acidic (negatively charged at pH 7)
Lys	Basic (positively charged at pH 7)
Ala	Non-polar (hydrophobic)
Cys	Polar (contains thiol group)
Pro	Non-polar (imino acid)
Phe	Non-polar (aromatic)

(a) Disulfide bridge: Cys (cysteine). Two cysteine residues can form a covalent disulfide bond ($-\mathrm{S}-\mathrm{S}-$) through oxidation of their thiol groups, stabilising tertiary and quaternary structure.

(b) Ionic bond: Glu and Lys. The carboxylate group of Glu ($-\mathrm{COO}^-$) and the amino group of Lys ($-\mathrm{NH}_3^+$) can form an electrostatic (ionic) interaction at physiological pH.

(c) Hydrogen bonds with water: Glu, Lys, and Cys. These amino acids have polar or charged side chains that can form hydrogen bonds with water, contributing to protein solubility. The non-polar residues (Val, Ala, Pro, Phe) tend to be buried in the protein interior, away from water.

Worked Example: Transcription and Translation -- Determining Gene Length

A protein consists of $400$ amino acids. The coding region of the gene includes $5$ exons and $4$ introns. The average intron length is $800\;\mathrm{bp}$ and the average exon length is $300\;\mathrm{bp}$. Calculate the length of the mature mRNA (excluding $5'$ cap and $3'$ poly-A tail) and the total length of the pre-mRNA.

Solution

Mature mRNA length (exons only): $5 \times 300 = 1500\;\mathrm{bp}$

Verification: a protein of $400$ amino acids requires $400 \times 3 = 1200\;\mathrm{bp}$ of coding sequence. The additional $300\;\mathrm{bp}$ in the exons are the $5'$ and $3'$ untranslated regions (UTRs) that are transcribed but not translated.

Pre-mRNA length (exons + introns): $1500 + (4 \times 800) = 1500 + 3200 = 4700\;\mathrm{bp}$

The introns constitute $\frac{3200}{4700} \approx 68\%$ of the pre-mRNA. This is typical of eukaryotic genes, which often have more intronic DNA than coding DNA. In humans, the average gene spans approximately $27000\;\mathrm{bp}$ but the average coding sequence is only about $1400\;\mathrm{bp}$.

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Common Pitfalls (Expanded)

• Confusing the sense and template strands: the template strand is the one used for transcription ($3' \to 5'$); the sense (coding) strand has the same sequence as the mRNA (with T instead of U).
• Writing DNA replication in the 3' to 5' direction: DNA polymerase synthesises exclusively in the $5' \to 3'$ direction. The template is read $3' \to 5'$, but the new strand is built $5' \to 3'$.
• Stating that "enzymes are used up in reactions": enzymes are catalysts that lower activation energy and are regenerated at the end of each catalytic cycle.
• Confusing alpha and beta glucose: the position of the OH group on C1 distinguishes them. Alpha-glucose polymers (starch, glycogen) are coiled and digestible; beta-glucose polymers (cellulose) are straight, form microfibrils, and are indigestible by most animals.
• Equating introns with "junk DNA": introns can contain regulatory sequences (enhancers, silencers), and are involved in alternative splicing, which allows one gene to produce multiple protein isoforms.
• Confusing saturated and unsaturated fatty acids: saturated fatty acids have no C=C double bonds and are straight (solid at room temperature); unsaturated fatty acids have C=C bonds creating kinks (liquid at room temperature).
• Stating that the genetic code is "universal without exception": the code is nearly universal but there are minor variations in mitochondrial DNA, some protozoa, and mycoplasma (e.g., UGA codes for tryptophan in mitochondria instead of being a stop codon).

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Exam-Style Problems

Problem 1: Data Analysis -- Enzyme Activity and Temperature

The following data show the rate of catalase activity (measured as $\mathrm{O}_2$ produced in $\mathrm{\mu mol/min}$) at different temperatures:

Temperature ($^\circ\mathrm{C}$)	Rate ($\mathrm{\mu mol/min}$)
10	8
20	18
30	32
37	40
45	25
55	5
65	0

(a) Plot the data and identify the optimum temperature. (b) Explain the molecular basis for the increase in rate from $10$ to $37^\circ\mathrm{C}$ and the decrease from $37$ to $65^\circ\mathrm{C}$. (c) A student claims that catalase from a thermophilic bacterium would show the same optimum temperature. Evaluate this claim.

Problem 2: Extended Response -- DNA Replication Fidelity

DNA polymerase has a proofreading ($3' \to 5'$ exonuclease) activity that reduces the error rate to approximately $10^{-9}$ per base pair per replication. (a) Explain how proofreading works at the molecular level. (b) Calculate the expected number of errors per replication of the human genome ($3.2 \times 10^9\;\mathrm{bp}$) with and without proofreading (error rate without proofreading $\approx 10^{-5}$). (c) Describe the role of mismatch repair enzymes in correcting errors that escape proofreading.

Problem 3: Quantitative -- Carbohydrate Identification Tests

A student performs biochemical tests on four unknown solutions (W, X, Y, Z) and records the following results: (i) Benedict's test: W = brick-red, X = blue, Y = blue (after acid hydrolysis: brick-red), Z = blue; (ii) Iodine test: W = blue-black, X = yellow-brown, Y = orange-brown, Z = blue-black; (iii) Biuret test: all negative. Identify the most likely carbohydrate in each solution and explain your reasoning.

Problem 4: Extended Response -- Protein Denaturation

A biochemist studies the effect of pH on the enzyme pepsin (stomach protease, optimum pH $\approx 2.0$) and salivary amylase (optimum pH $\approx 6.8$). (a) Explain why each enzyme has a different optimum pH, referring to the role of amino acid R groups in the active site. (b) Predict the effect of moving pepsin from pH $2.0$ to pH $7.0$ on its tertiary structure and activity. (c) Explain why denaturation by pH change is often reversible, whereas denaturation by high temperature is often irreversible.

Problem 5: Extended Response -- Central Dogma and One Gene-One Polypeptide

The central dogma of molecular biology states: DNA $\to$ RNA $\to$ Protein. (a) Describe the processes of transcription and translation, indicating where each occurs in a eukaryotic cell. (b) Evaluate the "one gene -- one polypeptide" hypothesis in light of alternative splicing and post-translational modifications. Provide a specific example of how one gene can produce multiple protein products. (c) Identify two exceptions to the central dogma (e.g., reverse transcription) and name the enzymes involved.

Problem 6: Data Analysis -- Chargaff's Rules and DNA Composition

Four DNA samples from different species are analysed for base composition:

Species	A (%)	T (%)	G (%)	C (%)
Human	30.4	30.1	19.6	19.9
E. coli	24.7	23.6	26.0	25.7
Wheat	27.3	27.1	22.7	22.8
Sample X	33.2	19.8	17.2	29.8

(a) Verify Chargaff's rules for the first three species. (b) Identify the anomaly in Sample X and explain what this suggests about the DNA in this sample. (c) Explain why the GC content varies between species and discuss its relevance to DNA melting temperature.

Problem 7: Extended Response -- Enzyme Inhibition and Drug Design

Statins are drugs that competitively inhibit HMG-CoA reductase, the rate-limiting enzyme in cholesterol biosynthesis. (a) Explain the mechanism of competitive inhibition and predict the effect of increasing dietary cholesterol on the efficacy of statins. (b) Describe how structural similarities between the statin molecule and the enzyme's natural substrate enable competitive inhibition. (c) Explain why statins are typically taken at night, relating this to the diurnal rhythm of cholesterol synthesis.

Problem 8: Quantitative -- Water Properties Calculation

The specific heat capacity of water is $4.18\;\mathrm{J/(g \cdot ^\circ C)}$. A $75\;\mathrm{kg}$ runner produces heat at a rate of $1200\;\mathrm{W}$ during exercise. Assuming $80\%$ of this heat is removed by sweating (latent heat of vaporisation of water $= 2260\;\mathrm{J/g}$) and the remaining $20\%$ is absorbed by body tissues (specific heat capacity of body tissue $\approx$ $3.5\;\mathrm{J/(g \cdot ^\circ C)}$), calculate: (a) the mass of water evaporated per minute, and (b) the rate of body temperature increase if sweating were completely prevented.

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If You Get These Wrong, Revise:

• Cell division and chromosome structure --> Review ./cell-biology
• Enzymes in physiological contexts --> Review ./human-physiology
• Genetic inheritance and gene expression --> Review ./genetics
• Plant biochemistry and photosynthesis --> Review ./plant-biology
• Evolution and molecular evidence --> Review ./ecology

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10. Water Properties in Biological Systems (Extended)

Water as a Temperature Buffer

The high specific heat capacity of water ($4.18\;\mathrm{J\;g^{-1}\;^\circ C^{-1}}$) is due to hydrogen bonding: energy input must overcome hydrogen bonds before the kinetic energy (and thus temperature) of water molecules can increase.

Biological significance:

• Aquatic environments have relatively stable temperatures (oceans vary by only a few degrees annually), providing a stable habitat for marine organisms.
• Organisms with high water content (most organisms) resist rapid temperature changes, buffering against thermal stress.
• Evaporative cooling (sweating in mammals, panting in dogs, transpiration in plants) dissipates large amounts of heat due to water's high latent heat of vaporisation ($2260\;\mathrm{J/g}$).

Water as a Solvent

Water's polarity makes it an excellent solvent for ionic compounds (e.g., $\mathrm{NaCl}$, $\mathrm{KCl}$) and polar molecules (sugars, amino acids, many proteins). This enables:

• Metabolic reactions in the cytoplasm (reactants dissolved in aqueous solution).
• Transport of dissolved substances in blood (glucose, amino acids, urea, hormones, ions).
• Nutrient uptake from soil by root hairs (mineral ions dissolved in soil water).

Non-polar molecules (lipids, steroid hormones, $\mathrm{O}_2$, $\mathrm{CO}_2$) are poorly soluble in water and require transport proteins or carriers (e.g., lipoproteins for cholesterol).

Water Density and Ice Formation

Water is one of the few substances that is less dense as a solid than as a liquid. This is because hydrogen bonds in ice form a regular hexagonal lattice with more space between molecules than in liquid water.

Significance: ice floats on water, insulating the liquid below and allowing aquatic organisms to survive beneath frozen surfaces in winter. If ice sank, lakes and oceans would freeze from the bottom up, killing most aquatic life.

Osmotic Quantitative Problems

Osmotic pressure ($\Pi$) can be calculated using the van't Hoff equation:

$\Pi = iCRT$

where $i$ is the van't Hoff factor (number of particles per formula unit), $C$ is molar concentration, $R$ is the gas constant ($0.0831\;\mathrm{L\;bar\;K^{-1}\;mol^{-1}}$), and $T$ is temperature in Kelvin.

For $\mathrm{NaCl}$ ($i = 2$, dissociates into $\mathrm{Na}^+$ and $\mathrm{Cl}^-$): $\Pi = 2 \times C \times 0.0831 \times T$

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11. Lipids in Detail (Extended)

Phospholipid Diversity

Phospholipids vary in their fatty acid chains and head groups:

Fatty acid variation:

• Chain length ($12$--$24$ carbons): shorter chains increase membrane fluidity.
• Degree of saturation: more unsaturated bonds = more fluidity (kinks prevent tight packing).
• Position of double bonds: omega-3 ($n$-3) and omega-6 ($n$-6) fatty acids are essential (must be obtained from the diet).

Head group variation:

Head group	Phospholipid name	Function
Choline	Phosphatidylcholine (lecithin)	Major component of eukaryotic membranes; surfactant in lungs.
Ethanolamine	Phosphatidylethanolamine	Abundant in bacterial membranes.
Serine	Phosphatidylserine	Normally on the inner leaflet; externalised during apoptosis (eat-me signal).
Inositol	Phosphatidylinositol	Precursor for IP$_3$ and DAG second messengers; PI 4,5-bisphosphate (PIP$_2$).
Glycerol	Phosphatidylglycerol	Cardiolipin in mitochondrial inner membrane.

Steroid Hormones

Cholesterol is the precursor for all steroid hormones:

Hormone	Source	Function
Testosterone	Testes (Leydig cells)	Male secondary sex characteristics; muscle growth; spermatogenesis.
Oestradiol	Ovaries (follicle)	Female secondary sex characteristics; menstrual cycle regulation; bone density.
Progesterone	Corpus luteum	Maintains endometrium during pregnancy; suppresses uterine contractions.
Cortisol	Adrenal cortex	Stress response; raises blood glucose; suppresses immune system.
Aldosterone	Adrenal cortex	$\mathrm{Na}^+$ reabsorption in kidneys; regulates blood pressure.
1,25-dihydroxyvitamin D	Kidney (with PTH)	Calcium absorption in gut; bone mineralisation.

Eicosanoids

Derived from arachidonic acid (a 20-carbon polyunsaturated fatty acid) by the action of cyclooxygenase (COX) or lipoxygenase (LOX) enzymes:

• Prostaglandins: diverse roles in inflammation, pain, fever, blood clotting, and gastric mucosa protection. COX inhibitors (aspirin, ibuprofen) reduce prostaglandin synthesis.
• Thromboxanes: promote platelet aggregation and vasoconstriction (thromboxane A2).
• Leukotrienes: involved in allergic and inflammatory responses (asthma, anaphylaxis).

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12. Protein Chemistry (Extended)

Amino Acid Properties

Amino acids are classified by the properties of their R groups:

Category	Amino acids	Properties
Non-polar (hydrophobic)	Gly, Ala, Val, Leu, Ile, Met, Phe, Trp, Pro	Tend to be buried in protein interior; drive protein folding via hydrophobic effect.
Polar, uncharged	Ser, Thr, Cys, Tyr, Asn, Gln	Form hydrogen bonds with water and other polar groups; often found on protein surface.
Acidic (negatively charged at pH 7)	Asp, Glu	Carboxylate groups ($-\mathrm{COO}^-$); often involved in salt bridges and catalytic sites.
Basic (positively charged at pH 7)	Lys, Arg, His	Amino groups ($-\mathrm{NH}_3^+$); bind to DNA/RNA; catalytic in enzyme active sites.

Protein Folding and the Hydrophobic Effect

The hydrophobic effect is the primary driving force for protein folding: non-polar side chains cluster together in the protein interior, minimising contact with water and maximising water-water hydrogen bonding (entropy-driven).

Anfinsen's dogma (1973): the native (functional) 3D structure of a protein is determined solely by its amino acid sequence (primary structure). This was demonstrated by the renaturation of ribonuclease A after denaturation with urea and $\beta$-mercaptoethanol.

Haemoglobin (Extended)

Structure: tetramer of $2\alpha$ and $2\beta$ globin chains, each with a haem group (protoporphyrin IX ring + $\mathrm{Fe}^{2+}$ ion). Each haem binds one $\mathrm{O}_2$ molecule, so each haemoglobin carries up to $4\;\mathrm{O}_2$.

Cooperative binding (sigmoidal curve): the binding of the first $\mathrm{O}_2$ molecule increases the affinity of the remaining haem groups for subsequent $\mathrm{O}_2$ molecules (positive cooperativity). This is described by the Monod-Wyman-Changeux (MWC) model or the sequential (KNF) model.

Bohr effect: increased $\mathrm{CO}_2$, decreased pH (increased $\mathrm{H}^+$), and increased temperature shift the oxygen dissociation curve to the right, reducing $\mathrm{Hb}$ affinity and promoting $\mathrm{O}_2$ unloading in metabolically active tissues.

$\mathrm{H}^+ + \mathrm{HbO}_2 \rightleftharpoons \mathrm{HHb} + \mathrm{O}_2$

2,3-BPG (bisphosphoglycerate): binds to deoxyhaemoglobin (T state), stabilising it and reducing $\mathrm{O}_2$ affinity. Concentrations increase at high altitude (adaptive response), enhancing $\mathrm{O}_2$ delivery to tissues.

Fetal haemoglobin (HbF): has two $\gamma$ chains instead of $\beta$ chains. Higher affinity for $\mathrm{O}_2$ than adult HbA (curve shifted left), facilitating $\mathrm{O}_2$ transfer from maternal to fetal blood across the placenta.

Collagen

The most abundant protein in mammals, providing tensile strength to connective tissues (skin, bone, tendon, cartilage).

Structure: a right-handed triple helix of three polypeptide chains (each called an $\alpha$-chain) with a repeating Gly-X-Y sequence (where X is often proline and Y is often hydroxyproline). Glycine is required at every third position because its small side chain allows the tight packing of the three chains.

Hydroxyproline and hydroxylysine are formed by post-translational modification (prolyl and lysyl hydroxylases, requiring vitamin C as a cofactor). These modifications stabilise the triple helix via hydrogen bonding. Vitamin C deficiency (scurvy) results in defective collagen and weakened connective tissues.

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13. Nucleic Acid Chemistry (Extended)

DNA Supercoiling

In addition to the double helix, DNA in cells is further twisted (supercoiled):

• Negative supercoiling: the DNA helix is underwound relative to relaxed B-DNA. Facilitates strand separation during transcription and replication.
• Positive supercoiling: the DNA helix is overwound; occurs ahead of the replication fork and transcription bubble, creating torsional strain.
• Topoisomerases relieve supercoiling by cutting one or both strands, allowing rotation, and resealing.

DNA Denaturation and Renaturation

Denaturation (melting): heating DNA separates the two strands by breaking hydrogen bonds. The melting temperature ($T_m$) is the temperature at which $50\%$ of the DNA is denatured.

$T_m = 69.3 + 0.41(\%\mathrm{GC}) - 0.72(\% \text{formamide})$

where $\%\mathrm{GC}$ is the percentage of G+C base pairs. Higher GC content $\to$ higher $T_m$ (because G-C pairs have 3 hydrogen bonds vs 2 for A-T pairs).

Renaturation (annealing): when cooled slowly, complementary strands find each other and re-form hydrogen bonds. The rate of renaturation depends on DNA concentration and complexity: highly repetitive sequences reanneal fastest; unique sequences slowest.

RNA Diversity

RNA type	Length	Function
mRNA	$500$--$10000\;\mathrm{nt}$	Carries genetic information from DNA to ribosome for translation.
tRNA	$\approx 75\;\mathrm{nt}$	Transports amino acids to the ribosome; anticodon recognises codon.
rRNA	$120$--$4700\;\mathrm{nt}$	Structural and catalytic component of ribosomes.
snRNA	$\approx 150\;\mathrm{nt}$	Components of the spliceosome; involved in pre-mRNA splicing.
miRNA	$\approx 22\;\mathrm{nt}$	Post-transcriptional regulation; binds to complementary sequences in target mRNA, leading to degradation or translational repression.
siRNA	$\approx 21\;\mathrm{nt}$	RNA interference (RNAi); triggers degradation of complementary mRNA.

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Exam-Style Problems (Extended)

Problem 9: Extended Response -- Protein Structure and Enzyme Function

Lysozyme is an enzyme found in tears and egg white that hydrolyses the $\beta$-1,4-glycosidic bonds in bacterial peptidoglycan. (a) Explain how the tertiary and quaternary structure of lysozyme contributes to its function, referring to the roles of specific amino acid residues in the active site. (b) Explain why lysozyme is effective against Gram-positive bacteria but less so against Gram-negative bacteria. (c) If the pH of a solution containing lysozyme is raised from 5.0 to 8.0, predict the effect on enzyme activity and explain the molecular basis for this change.

Problem 10: Data Analysis -- Water Potential and Osmosis

Four potato cylinders of identical dimensions are placed in sucrose solutions of different concentrations. After 2 hours, the mass change of each cylinder is measured. The water potential of pure water is $0\;\mathrm{kPa}$. The relationship between sucrose concentration and water potential is approximately $\psi_s = -iCRT$ where $i = 1$ for sucrose. (a) If a potato cylinder in $0.3\;\mathrm{mol/L}$ sucrose shows no mass change, estimate the water potential of the potato tissue. (b) Predict the direction of water movement in $0.1\;\mathrm{mol/L}$ and $0.5\;\mathrm{mol/L}$ sucrose. (c) Calculate the pressure potential of the cells in $0.1\;\mathrm{mol/L}$ sucrose at equilibrium, given that the solute potential is $-249\;\mathrm{kPa}$.

Problem 11: Quantitative -- Haemoglobin and Oxygen Transport

At sea level, alveolar $\mathrm{pO}_2 = 13.3\;\mathrm{kPa}$ and venous $\mathrm{pO}_2 = 5.3\;\mathrm{kPa}$. Using a standard oxygen dissociation curve: (a) estimate the percentage saturation of haemoglobin in arterial blood and venous blood. (b) Calculate the amount of $\mathrm{O}_2$ unloaded per $100\;\mathrm{mL}$ of blood, assuming each gram of haemoglobin can carry $1.34\;\mathrm{mL}$ of $\mathrm{O}_2$ and blood contains $150\;\mathrm{g/L}$ of haemoglobin. (c) At high altitude ($5000\;\mathrm{m}$), alveolar $\mathrm{pO}_2$ drops to $8.0\;\mathrm{kPa}$. Predict the effect on oxygen loading and explain the compensatory mechanisms (polycythaemia, increased 2,3-BPG).

Problem 12: Extended Response -- Lipids and Health

A patient has a blood cholesterol level of $280\;\mathrm{mg/dL}$ (desirable: $< 200\;\mathrm{mg/dL}$). LDL cholesterol is $190\;\mathrm{mg/dL}$ and HDL cholesterol is $35\;\mathrm{mg/dL}$. (a) Explain the roles of LDL and HDL in cholesterol transport. (b) Calculate the total cholesterol:HDL ratio and assess the cardiovascular risk. (c) Explain the mechanism of action of statins in lowering blood cholesterol. (d) Discuss why saturated fat intake increases LDL cholesterol and why omega-3 fatty acids may be beneficial.

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Additional Worked Examples

Worked Example: Protein Structure and Amino Acid Properties

A protein has the following amino acid sequence at its N-terminus: Met-Ala-Glu-Lys-Leu-Val-Pro-Phe-Arg-Ser-Cys

(a) Identify the amino acids that are: hydrophobic, hydrophilic, acidic, basic, and those that can form disulphide bonds. (b) At pH 7.4, which amino acids will be positively charged and which will be negatively charged? (c) Predict the likely secondary structure of the sequence Ala-Leu-Val-Pro-Phe. (d) Explain how the sequence Glu-Lys might affect protein folding through ionic interactions.

Solution

(a) Classification:

• Hydrophobic: Ala, Leu, Val, Phe (non-polar R groups)
• Hydrophilic/polar: Ser, Glu, Lys, Arg (polar or charged R groups)
• Acidic (negatively charged at neutral pH): Glu (glutamic acid, $\mathrm{p}K_a \approx 4.3$)
• Basic (positively charged at neutral pH): Lys (lysine, $\mathrm{p}K_a \approx 10.5$), Arg (arginine, $\mathrm{p}K_a \approx 12.5$)
• Disulphide bond formation: Cys (cysteine) -- two Cys residues can form a covalent disulphide bridge ($-\mathrm{S}-\mathrm{S}-$)
• Structural/disruptive: Pro (proline) -- its cyclic R group creates a rigid kink in the polypeptide chain
• Special: Met (methionine) -- moderately hydrophobic, can be the start amino acid

(b) At pH 7.4:

• Positively charged: Lys (side chain $\mathrm{NH}_3^+$), Arg (side chain guanidinium $\mathrm{NH}_2^+$), and the N-terminal amino group ($\mathrm{NH}_3^+$).
• Negatively charged: Glu (side chain $\mathrm{COO}^-$), and the C-terminal carboxyl group ($\mathrm{COO}^-$).
• Neutral: Ala, Leu, Val, Pro, Phe, Ser, Cys, Met.

(c) The sequence Ala-Leu-Val-Phe consists of small and hydrophobic amino acids (except for the Pro disruption). Without Pro, this would be a strong candidate for an alpha helix (hydrophobic amino acids often form helical regions in globular proteins, especially in the core). However, Pro is a known helix breaker because its rigid cyclic structure cannot adopt the phi/psi angles required for alpha-helical conformation. The Ala-Leu-Val segment may form part of a helix, but Pro terminates it.

(d) At pH 7.4, Glu is negatively charged ($\mathrm{COO}^-$) and Lys is positively charged ($\mathrm{NH}_3^+$). These two residues, if close in the folded protein (even if distant in the primary sequence), can form an ionic bond (salt bridge): the negatively charged carboxylate of Glu is attracted to the positively charged ammonium of Lys. Salt bridges stabilise the tertiary structure of proteins and are particularly important in extremophilic proteins (e.g., thermophilic enzymes have more salt bridges for heat stability).

Worked Example: Enzyme Catalysis and Activation Energy

An enzyme-catalysed reaction has an activation energy ($E_a$) of $35\;\mathrm{kJ/mol}$. The uncatalysed reaction has an activation energy of $85\;\mathrm{kJ/mol}$. The reaction is carried out at $310\;\mathrm{K}$ ($37^\circ\mathrm{C}$). (a) Calculate the rate enhancement factor using the Arrhenius equation ($k = Ae^{-E_a/RT}$, where $R = 8.314\;\mathrm{J/(mol \cdot K)}$). (b) If the temperature is increased by $10^\circ\mathrm{C}$ ($320\;\mathrm{K}$), calculate the new rate enhancement. (c) Explain why the rate enhancement is a ratio, not an absolute value. (d) Explain how the enzyme lowers the activation energy.

Solution

(a) Rate ratio $= \frac{k_{catalysed}}{k_{uncatalysed}} = \frac{Ae^{-E_{a(cat)}/RT}}{Ae^{-E_{a(uncat)}/RT}}$ (assuming the pre-exponential factor $A$ is the same for both) $= e^{-(E_{a(cat)} - E_{a(uncat)})/RT}$ $= e^{-(35000 - 85000)/(8.314 \times 310)}$ $= e^{-(-50000)/2577}$ $= e^{19.40}$ $= 2.62 \times 10^8$

The enzyme increases the reaction rate by approximately $262$ million times.

(b) At $320\;\mathrm{K}$: $= e^{50000/(8.314 \times 320)}$ $= e^{50000/2660}$ $= e^{18.80}$ $= 1.47 \times 10^8$

The rate enhancement is slightly lower at the higher temperature ($147$ million vs $262$ million times). This is because the uncatalysed reaction also speeds up with temperature (exponential dependence on $T$), partially closing the gap.

(c) The rate enhancement is a ratio because it represents how many times faster the enzyme makes the reaction compared to no enzyme, at the same temperature. It is dimensionless and independent of the absolute reaction rate, the enzyme concentration, and the substrate concentration (as long as the enzyme is operating under $V_{max}$ conditions).

(d) The enzyme lowers the activation energy by:

1. Binding the substrate in the active site, creating an enzyme-substrate (ES) complex. The enzyme holds the substrate in a conformation that strains bonds (distorting the substrate toward the transition state geometry).
2. Stabilising the transition state: the active site is complementary to the transition state (not the substrate). Binding energy released when the transition state is formed lowers the energy barrier.
3. Providing a favourable microenvironment: the active site may exclude water (reducing dielectric constant), orient catalytic groups precisely, or provide acidic/basic residues that participate directly in the reaction (acid-base catalysis, covalent catalysis).
4. Induced fit: the enzyme changes conformation upon substrate binding, optimising the active site geometry for catalysis.

Worked Example: DNA and RNA Comparison

(a) A double-stranded DNA molecule has $2400$ base pairs and is $42\%$ GC. Calculate the number of hydrogen bonds and the length of the DNA molecule. (b) If this DNA is transcribed into mRNA and the mRNA is $2100$ nucleotides long, how many introns were present and what was the total length of the introns? (c) The mRNA is translated. How many amino acids are in the resulting protein (excluding the stop codon)? (d) Explain why RNA is less stable than DNA and how this affects its biological function.

Solution

(a) Total base pairs $= 2400$. Total nucleotides $= 4800$. GC content $= 42\%$: $\mathrm{G} + \mathrm{C} = 0.42 \times 4800 = 2016$ nucleotides. $\mathrm{G} = \mathrm{C} = 1008$ nucleotides each. $\mathrm{A} + \mathrm{T} = 0.58 \times 4800 = 2784$ nucleotides. $\mathrm{A} = \mathrm{T} = 1392$ nucleotides each.

Hydrogen bonds: $\mathrm{G} \equiv \mathrm{C}$ (3 H-bonds) and $\mathrm{A} = \mathrm{T}$ (2 H-bonds). Total H-bonds $= 1008 \times 3 + 1392 \times 2 = 3024 + 2784 = 5808$ hydrogen bonds.

Length: $0.34\;\mathrm{nm}$ per base pair. Total length $= 2400 \times 0.34 = 816\;\mathrm{nm} = 0.816\;\mathrm{\mu m}$.

(b) Coding sequence: $2100$ nucleotides (mature mRNA). Pre-mRNA length: $2400 \times 2$ (both strands are transcribed? No -- only one strand is transcribed). Actually, the DNA coding region is $2400\;\mathrm{bp}$ on one strand. But not all of the DNA may be transcribed. If the gene spans $2400\;\mathrm{bp}$ of DNA (template strand), the primary transcript (pre-mRNA) would be approximately $2400\;\mathrm{nt}$ long (ignoring the promoter and terminator regions beyond the gene).

Wait -- the pre-mRNA includes both exons and introns. If the gene is $2400\;\mathrm{bp}$ of transcribed DNA: Pre-mRNA $= 2400\;\mathrm{nt}$. Mature mRNA $= 2100\;\mathrm{nt}$. Total intron length $= 2400 - 2100 = 300\;\mathrm{nt}$.

But this assumes the gene is exactly $2400\;\mathrm{bp}$. The question does not specify this directly. If the DNA is $2400\;\mathrm{bp}$ total and the mRNA is $2100\;\mathrm{nt}$, the introns account for $300\;\mathrm{nt}$. The number of introns cannot be determined without additional information (e.g., knowing the exon sizes).

(c) Amino acids: $2100 / 3 = 700$ codons. If the last codon is a stop codon, the protein has $699$ amino acids.

(d) RNA is less stable than DNA because:

• RNA contains ribose (with a $2'$-$\mathrm{OH}$ group) instead of deoxyribose. The $2'$-$\mathrm{OH}$ makes RNA susceptible to alkaline hydrolysis (it can act as a nucleophile, attacking the phosphodiester bond).
• RNA is typically single-stranded, so it has no complementary strand for repair.
• Uracil in RNA (instead of thymine) makes it harder to detect and repair deamination damage.
• These properties make RNA suitable as a transient informational molecule (mRNA) and catalytic molecule (ribozymes, rRNA, tRNA), but unsuitable for long-term genetic storage. DNA's stability (deoxyribose, double-stranded, thymine) makes it ideal as the permanent genetic material.

Worked Example: Competitive and Non-Competitive Inhibition Graphically

An enzyme has $K_m = 5.0\;\mathrm{mM}$ and $V_{max} = 100\;\mathrm{\mu mol/min}$. A competitive inhibitor at $2.0\;\mathrm{mM}$ gives $K_m^{app} = 15.0\;\mathrm{mM}$, $V_{max}^{app} = 100$. A non-competitive inhibitor at $5.0\;\mathrm{mM}$ gives $K_m^{app} = 5.0\;\mathrm{mM}$, $V_{max}^{app} = 50$. (a) Sketch the Michaelis-Menten curves for all three conditions. (b) Calculate the initial velocity ($v$) at $[S] = 5.0\;\mathrm{mM}$ for each condition. (c) Calculate the substrate concentration needed to reach $50\%$ of $V_{max}$ in each condition.

Solution

(a) [Students should sketch $v$ vs $[S]$ curves:]

• No inhibitor: hyperbolic curve reaching $V_{max} = 100$ at high $[S]$, half-maximal at $[S] = K_m = 5.0$.
• Competitive: curve reaches the same $V_{max} = 100$ but is shifted to the right (half-maximal at $[S] = K_m^{app} = 15.0$). The initial slope is lower.
• Non-competitive: curve reaches a lower $V_{max}^{app} = 50$ but has the same half-maximal point ($[S] = K_m = 5.0$). The curve is lower at all substrate concentrations.

(b) $v = \frac{V_{max}[S]}{K_m + [S]}$:

No inhibitor: $v = \frac{100 \times 5.0}{5.0 + 5.0} = 50\;\mathrm{\mu mol/min}$

Competitive: $v = \frac{100 \times 5.0}{15.0 + 5.0} = \frac{500}{20} = 25\;\mathrm{\mu mol/min}$

Non-competitive: $v = \frac{50 \times 5.0}{5.0 + 5.0} = \frac{250}{10} = 25\;\mathrm{\mu mol/min}$

At $[S] = K_m$, both inhibitors reduce the velocity by the same amount ($50\%$), but by different mechanisms.

(c) $50\%$ of $V_{max}$ is achieved when $[S] = K_m$ (by definition).

• No inhibitor: $[S] = 5.0\;\mathrm{mM}$
• Competitive: $[S] = K_m^{app} = 15.0\;\mathrm{mM}$
• Non-competitive: $[S] = K_m^{app} = 5.0\;\mathrm{mM}$ (same $K_m$, but $50\%$ of the reduced $V_{max}^{app} = 25$, achieved at $[S] = 5.0\;\mathrm{mM}$).

For the non-competitive inhibitor, to reach $50\%$ of the original $V_{max}$ ($= 50\;\mathrm{\mu mol/min}$): $v = \frac{50[S]}{5 + [S]} = 50$, which requires $[S] \to \infty$ (the enzyme can never reach the original $V_{max}$). This illustrates a key difference: competitive inhibition can be overcome by increasing $[S]$, but non-competitive cannot.

Worked Example: Water Properties and Biological Significance

(a) Explain why water has a high specific heat capacity ($4.18\;\mathrm{J/(g \cdot ^\circ C)}$) and why this is important for organisms. (b) Calculate the energy required to raise the temperature of $70\;\mathrm{kg}$ of water (human body) by $1^\circ\mathrm{C}$. (c) Explain why ice floats on water and why this is ecologically significant. (d) Explain the role of water as a solvent for biological molecules, distinguishing between hydrophilic and hydrophobic substances.

Solution

(a) Water has a high specific heat capacity because of hydrogen bonding: energy input must first break hydrogen bonds between water molecules before the kinetic energy (temperature) of individual molecules can increase. This means water absorbs a large amount of heat energy with only a small temperature change. Biological significance: water buffers temperature changes in organisms and environments (oceans, large lakes), providing thermal stability for biochemical reactions.

(b) $Q = mc\Delta T = 70000\;\mathrm{g} \times 4.18\;\mathrm{J/(g \cdot ^\circ C)} \times 1^\circ\mathrm{C} = 292\,600\;\mathrm{J} = 292.6\;\mathrm{kJ}$.

(c) Ice floats because it is less dense than liquid water. In ice, each water molecule forms 4 hydrogen bonds in a rigid, open hexagonal lattice that is less densely packed than the disordered arrangement in liquid water. Ecological significance: the layer of ice on the surface of lakes and ponds insulates the water below, preventing it from freezing solid and allowing aquatic organisms to survive winter. If ice sank, water bodies would freeze from the bottom up, killing most aquatic life.

(d) Water is an excellent solvent for polar and ionic substances (hydrophilic) because its polar molecules can form hydrogen bonds and ion-dipole interactions with solutes. Examples: $\mathrm{Na}^+$, $\mathrm{Cl}^-$, glucose, amino acids, DNA, proteins dissolve in water.

Non-polar substances (hydrophobic) cannot form hydrogen bonds with water and are insoluble. Examples: lipids, cholesterol, steroid hormones. In aqueous environments, hydrophobic molecules aggregate (hydrophobic effect), which is the driving force for membrane formation (lipid bilayers) and protein folding (hydrophobic core).

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Additional Common Pitfalls

• Confusing alpha helices and beta sheets: alpha helices are spiral structures stabilised by hydrogen bonds between the C=O of residue $i$ and the N-H of residue $i+4$; beta sheets are extended strands stabilised by hydrogen bonds between adjacent strands (parallel or antiparallel).
• Confusing primary structure with amino acid sequence: they are the same thing. The primary structure IS the sequence of amino acids.
• Assuming all enzymes are proteins: ribozymes (e.g., rRNA in ribosomes, self-splicing introns) are RNA molecules with catalytic activity.
• Confusing DNA and RNA bases: DNA has A, T, G, C; RNA has A, U, G, C. Uracil replaces thymine.
• Forgetting that the genetic code is degenerate but not ambiguous: most amino acids are coded for by more than one codon (degenerate), but each codon codes for only one amino acid (not ambiguous).
• Confusing saturated and unsaturated fatty acids: saturated fats have no double bonds (straight chains, solid at room temperature); unsaturated fats have one or more C=C double bonds (kinked chains, liquid at room temperature).

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Additional Exam-Style Problems with Full Solutions

Problem 13: Extended Response -- Protein Denaturation and Renaturation

(a) Describe the four levels of protein structure, giving examples of the bonds and interactions that stabilise each level. (b) Explain how heat, pH changes, and heavy metals cause protein denaturation. (c) Explain why some proteins can renature after denaturation while others cannot. (d) Explain why denaturation of an enzyme abolishes its catalytic activity but does not destroy its primary structure.

Answer 13

(a) Primary structure: the sequence of amino acids, stabilised by peptide (covalent) bonds. Example: insulin A chain = 21 amino acids, B chain = 30 amino acids.

Secondary structure: local folding into regular structures (alpha helix, beta pleated sheet), stabilised by hydrogen bonds between backbone C=O and N-H groups (not R groups).

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

• Hydrophobic interactions (hydrophobic R groups cluster in the interior)
• Hydrogen bonds (between R groups)
• Ionic bonds (salt bridges between charged R groups)
• Disulphide bonds (covalent bonds between Cys residues)
• Van der Waals forces (between closely packed atoms)

Quaternary structure: the arrangement of multiple polypeptide subunits into a functional protein, stabilised by the same interactions as tertiary structure (but between subunits, not within a subunit). Example: haemoglobin ($\alpha_2\beta_2$, four subunits).

(b) Heat: increases kinetic energy, disrupting weak interactions (hydrogen bonds, hydrophobic interactions, ionic bonds). The protein unfolds, exposing hydrophobic R groups. At very high temperatures, peptide bonds may also break.

pH changes: alter the ionisation state of R groups. At extreme pH, acidic groups may be protonated (or basic groups deprotonated), disrupting ionic bonds and salt bridges. This changes the charge distribution, causing unfolding.

Heavy metals (e.g., $\mathrm{Hg}^{2+}$, $\mathrm{Pb}^{2+}$, $\mathrm{Ag}^+$): bind to sulfhydryl groups ($-\mathrm{SH}$) of cysteine residues, forming strong covalent or coordination bonds that disrupt disulphide bridges and the active site geometry of enzymes. Heavy metals can also bind to carboxyl and amino groups, disrupting ionic bonds.

(c) Renaturable proteins: small, single-domain proteins (e.g., ribonuclease A, as demonstrated by Anfinsen's experiment) can refold spontaneously after denaturation because the primary structure contains all the information needed for correct folding. The protein follows the thermodynamically most stable conformation (the native state has the lowest free energy).

Non-renaturable proteins: large, multi-domain proteins or those with disulphide bonds may not refold correctly because: (1) the denatured state has many possible conformations, and the correct one may not be found within a reasonable time (kinetic trapping); (2) disulphide bonds may form incorrectly; (3) aggregation may occur (exposed hydrophobic groups of multiple unfolded proteins stick together irreversibly); (4) chaperone proteins (normally present in the cell) are absent in vitro.

(d) Denaturation disrupts the secondary, tertiary, and quaternary structure of the enzyme, destroying the precise 3D arrangement of the active site. Without the correct active site geometry, the enzyme cannot bind its substrate or catalyse the reaction. However, denaturation does not break peptide bonds (the primary structure is preserved). If the denaturing agent is removed (and the protein is renaturable), the enzyme can regain its activity.

Problem 14: Quantitative -- Nucleic Acid Quantitation

A DNA sample has an absorbance at $260\;\mathrm{nm}$ ($A_{260}$) of $0.50$ in a $1\;\mathrm{cm}$ cuvette. The absorbance at $280\;\mathrm{nm}$ ($A_{280}$) is $0.28$. (a) Calculate the DNA concentration in $\mathrm{\mu g/mL}$ (using the conversion: $A_{260} = 1.0$ corresponds to $50\;\mathrm{\mu g/mL}$ double-stranded DNA). (b) Calculate the $A_{260}/A_{280}$ ratio and assess the purity of the sample. (c) If the sample is diluted $1:10$ before measurement, what was the original concentration? (d) Explain why protein contamination increases $A_{280}$.

Answer 14

(a) DNA concentration $= A_{260} \times 50\;\mathrm{\mu g/mL} = 0.50 \times 50 = 25\;\mathrm{\mu g/mL}$.

(b) $A_{260}/A_{280} = 0.50 / 0.28 = 1.79$. Pure DNA has a ratio of approximately $1.8$. A ratio of $1.79$ indicates good purity (minimal protein contamination). Ratios below $1.7$ suggest significant protein contamination; ratios above $2.0$ suggest RNA contamination or other contaminants.

(c) Original concentration $= 25 \times 10 = 250\;\mathrm{\mu g/mL}$.

(d) Proteins contain aromatic amino acids (tryptophan, tyrosine) that absorb UV light at $280\;\mathrm{nm}$ (with a weaker absorbance at $260\;\mathrm{nm}$). When protein is present as a contaminant in the DNA sample, it increases $A_{280}$ more than $A_{260}$, lowering the $A_{260}/A_{280}$ ratio. Pure DNA has minimal absorbance at $280\;\mathrm{nm}$ (DNA bases absorb maximally at $260\;\mathrm{nm}$).

Problem 15: Extended Response -- Carbohydrates: Structure and Function

Compare and contrast the structures and functions of: (a) starch (amylose and amylopectin), (b) cellulose, (c) glycogen, and (d) chitin. For each, describe: the type of glycosidic bond, whether it is alpha or beta, the degree of branching, the organisms that produce it, and its biological function. Explain why humans can digest starch and glycogen but not cellulose or chitin.

Answer 15

Feature	Starch (amylose)	Starch (amylopectin)	Glycogen	Cellulose	Chitin
Monomer	$\alpha$-glucose	$\alpha$-glucose	$\alpha$-glucose	$\beta$-glucose	N-acetylglucosamine
Bond	$\alpha$-1,4	$\alpha$-1,4 and $\alpha$-1,6	$\alpha$-1,4 and $\alpha$-1,6	$\beta$-1,4	$\beta$-1,4
Structure	Helical (spiral)	Branched helices	Highly branched	Linear, parallel chains (microfibrils)	Linear, parallel chains
Branching	None	Every 24--30 residues	Every 8--12 residues	None	None
Producers	Plants	Plants	Animals, fungi	Plants, algae	Fungi, arthropods, some animals
Function	Energy storage	Energy storage	Energy storage (rapid mobilisation)	Structural (cell walls)	Structural (exoskeletons, fungal walls)

Why humans can digest starch/glycogen but not cellulose/chitin:

Humans produce amylase (in saliva and pancreas) that hydrolyses $\alpha$-1,4 glycosidic bonds and isomaltase (brush border enzyme) that hydrolyses $\alpha$-1,6 bonds. These enzymes can access the glycosidic bonds in starch and glycogen because the $\alpha$ configuration creates a helical structure with bonds accessible to the enzyme active site.

Humans do not produce cellulase or chitinase, the enzymes needed to hydrolyse $\beta$-1,4 glycosidic bonds. The $\beta$ configuration creates linear, tightly packed chains (in cellulose) or chains with bulky N-acetyl groups (in chitin) that are inaccessible to human digestive enzymes. Cellulose and chitin pass through the digestive system as dietary fibre (insoluble fibre).

Ruminants (cows, sheep) and termites can digest cellulose because they harbour symbiotic microorganisms (bacteria, protozoa) in their digestive tracts that produce cellulase.

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Cross-References to Related Topics

• Enzyme kinetics in detail: Review ./metabolism-cell-biology for Michaelis-Menten kinetics, inhibition types, and enzyme regulation.
• DNA replication and transcription: Review ./genetics-advanced for detailed mechanisms of DNA and RNA synthesis.
• Cell membranes and lipids: Review ./cell-biology for membrane structure and the fluid mosaic model.
• Protein synthesis: Review ./genetics-advanced for translation and post-translational modifications.
• Immunology and antibody structure: Review ./immunology for immunoglobulin structure and antigen-antibody interactions.
• Plant biology and carbohydrates: Review ./plant-biology for cellulose, starch, and transport sugars.

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Supplementary: Molecular Biology Techniques (HL Extension)

Gel Electrophoresis Principles

Gel electrophoresis separates DNA, RNA, or protein molecules based on their size, charge, and shape by applying an electric field across a gel matrix.

DNA electrophoresis:

• DNA is negatively charged (phosphate backbone) and migrates toward the positive electrode (anode).
• Agarose gel: concentration $0.5$--$2.0\%$ (higher concentration = smaller pores = better separation of small fragments). Suitable for fragments $100$--$25\,000\;\mathrm{bp}$.
• Polyacrylamide gel (PAGE): higher resolution, suitable for fragments $1$--$1000\;\mathrm{bp}$.
• Migration distance is inversely proportional to the log of fragment size (larger fragments move more slowly through the gel matrix).
• A DNA ladder (fragments of known size) is run alongside samples for size estimation.

Visualisation: DNA is stained with ethidium bromide or SYBR Safe (intercalates between bases and fluoresces under UV light). Alternatively, samples can be loaded with loading dye (contains glycerol or sucrose to make the sample denser than the buffer, plus tracking dyes).

Polymerase Chain Reaction (PCR) -- Detailed Mechanism

PCR amplifies a specific DNA sequence exponentially. Components:

• Template DNA: the DNA containing the target sequence.
• Two primers: short, single-stranded DNA oligonucleotides (typically $18$--$25\;\mathrm{nt}$) complementary to the regions flanking the target. The $3'$ end of each primer faces the target sequence.
• Taq DNA polymerase: a thermostable DNA polymerase from Thermus aquaticus (optimum temperature $72^\circ\mathrm{C}$, withstands the $95^\circ\mathrm{C}$ denaturation step).
• dNTPs: deoxynucleoside triphosphates (dATP, dCTP, dGTP, dTTP) -- building blocks for DNA synthesis.
• Buffer: provides optimal pH and $\mathrm{Mg}^{2+}$ concentration (cofactor for Taq polymerase).

Three-step cycle (repeated $25$--$35$ times):

1. Denaturation ($94$--$98^\circ\mathrm{C}$, $30\;\mathrm{s}$): hydrogen bonds break, separating the double-stranded DNA into single strands.

2. Annealing ($50$--$65^\circ\mathrm{C}$, $30\;\mathrm{s}$): primers hybridise (bind) to their complementary sequences on the single-stranded template. The annealing temperature depends on the primer sequences (typically $3$--$5^\circ\mathrm{C}$ below the primer $T_m$).

3. Extension ($72^\circ\mathrm{C}$, $1\;\mathrm{min/kb}$): Taq polymerase extends the primers, synthesising new DNA strands complementary to the template.

Result: after $n$ cycles, $2^n$ copies of the target are produced (plus the original template).

Restriction Enzyme Analysis

Restriction enzymes recognise specific palindromic DNA sequences and cut both strands, producing fragments of defined lengths. Applications:

• DNA fingerprinting: VNTR (variable number tandem repeat) or STR (short tandem repeat) analysis. PCR amplifies multiple STR loci, and the fragment sizes are compared between samples.
• Diagnostic testing: detecting the presence of a disease-causing mutation by checking whether a restriction site is created or destroyed by the mutation (RFLP -- restriction fragment length polymorphism).
• Cloning: cutting both the insert and vector with the same enzyme(s) produces compatible ends for ligation.

DNA Sequencing

Sanger sequencing (chain termination method):

1. Four separate reactions are set up, each containing the template, primer, DNA polymerase, dNTPs, and a different ddNTP (dideoxynucleotide: lacks the $3'$-$\mathrm{OH}$ group needed for chain elongation).
2. When a ddNTP is incorporated, chain termination occurs. Each reaction produces a mixture of fragments of different lengths, each ending with the specific ddNTP.
3. Fragments are separated by capillary electrophoresis and detected by fluorescence (modern automated sequencers use fluorescently labelled ddNTPs, allowing all four reactions in one tube).

Next-generation sequencing (NGS): massively parallel sequencing technologies that sequence millions of DNA fragments simultaneously. Examples: Illumina (sequencing by synthesis), PacBio (single-molecule real-time sequencing), Oxford Nanopore (sequencing by measuring current changes as DNA passes through a protein nanopore). NGS has revolutionised genomics, enabling whole-genome sequencing in hours at rapidly decreasing cost.

Worked Example: Restriction Fragment Length Analysis

A gene has a single EcoRI site (GAATTC) in its normal allele but the site is mutated (GAATTC $\to$ GAA TTC, introducing a 1 bp insertion that destroys the site) in a disease allele. The gene is $3000\;\mathrm{bp}$ long. (a) How many fragments are produced by EcoRI digestion of each allele? (b) If the EcoRI site is at position $1500$, what are the fragment sizes for each allele? (c) How could this be used for genetic diagnosis?

Solution

(a) Normal allele: one EcoRI site produces 2 fragments. Disease allele: no EcoRI site (the site is destroyed by the mutation), producing 1 fragment (the uncut gene).

(b) Normal allele: fragments are $1500\;\mathrm{bp}$ and $1500\;\mathrm{bp}$ (EcoRI cuts at position $1500$). Disease allele: single fragment of $3001\;\mathrm{bp}$ (3000 bp gene + 1 bp insertion).

(c) Diagnostic use: extract genomic DNA from the patient, digest with EcoRI, run on agarose gel.

• If the patient is homozygous normal: 2 bands at $1500\;\mathrm{bp}$.
• If the patient is homozygous disease: 1 band at $3001\;\mathrm{bp}$.
• If the patient is heterozygous: 3 bands at $3001$, $1500$, and $1500\;\mathrm{bp}$.

This is a simple form of RFLP analysis. In practice, PCR amplification of the region followed by restriction digestion (PCR-RFLP) is more commonly used for diagnostic testing, as it requires less DNA and is faster.