Molecular Biology — IB Biology HL HL Study Guide

For: IB Biology HL candidates sitting IB Biology HL.

Covers: core molecular biology content including water biochemistry, 4 major biomolecule classes, DNA replication/transcription/translation, enzyme function, and photosynthesis/respiration overviews

You should already know: IGCSE Biology, basic chemistry.

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

1. What Is Molecular Biology?

Molecular biology is the study of the molecular basis of biological activity, exploring how biomolecules interact to carry out the core processes of life, from inheritance to energy metabolism. Aligned with IB DP Biology HL Topic 2, this content makes up ~15% of your total HL assessment marks, and is tested across Paper 1 (multiple choice), Paper 2 (structured response), and Paper 3 (data analysis and option topics). Common related terms you will see in exam questions include biochemical genetics, molecular life sciences, and metabolic biochemistry.

2. Water and biochemistry essentials

All biological reactions occur in aqueous environments, so water’s unique properties are the foundation of all biochemistry. Water is a polar molecule: covalent bonds between oxygen and hydrogen pull shared electrons toward the more electronegative oxygen, creating a partially negative oxygen end and partially positive hydrogen ends. This polarity allows hydrogen bonds (weak intermolecular attractions) to form between adjacent water molecules, which drive all of water’s core properties:

1. Thermal properties: High specific heat capacity ($4.2{\text{ J g}}^{-1}°{\text{C}}^{-1}$) stabilizes organism and ecosystem temperatures; high latent heat of vaporization ($2260{\text{ J g}}^{-1}$) makes sweat an effective coolant.
2. Cohesive/adhesive properties: Cohesion (attraction between water molecules) creates surface tension and supports continuous water columns in plant xylem; adhesion (attraction between water and polar surfaces) drives capillary action.
3. Solvent properties: Polar and ionic compounds dissolve easily in water, making it the primary transport medium for blood, cytoplasm, and plant vascular tissue.

Worked example: Explain why coastal cities have smaller temperature fluctuations between day and night than inland desert cities. [2 marks] Solution: Coastal areas are surrounded by large bodies of water, which have a very high specific heat capacity due to extensive hydrogen bonding between water molecules [1 mark]. This means water absorbs large amounts of heat during the day without a large temperature increase, and releases heat slowly at night, stabilizing air temperatures [1 mark]. Exam tip: You will lose 1 mark if you do not explicitly link water properties to hydrogen bonding in exam responses.

3. Carbohydrates, lipids, proteins, nucleic acids

The four core classes of carbon-based biomolecules make up all structural and functional components of living cells:

• Carbohydrates: Monomers are monosaccharides (e.g. glucose, ribose) with the general formula ${\text{C}}_n({\text{H}}_2\text{O}{)}_n$. Disaccharides (e.g. sucrose, maltose) form via glycosidic bonds between two monosaccharides; polysaccharides (starch, glycogen, cellulose) are long-chain polymers used for energy storage or structural support. Energy density = $17{\text{ kJ g}}^{-1}$.
• Lipids: Non-polar, hydrophobic molecules including triglycerides (3 fatty acids linked to glycerol via ester bonds), phospholipids (amphipathic molecules with hydrophilic phosphate heads and hydrophobic fatty acid tails, the core component of cell membranes), and steroids (e.g. cholesterol, sex hormones). Energy density = $37{\text{ kJ g}}^{-1}$, making them ideal for long-term energy storage.
• Proteins: Monomers are 20 distinct amino acids, linked via peptide bonds. Four levels of structure: primary (linear amino acid sequence), secondary (alpha helices or beta-pleated sheets held by hydrogen bonds), tertiary (3D folded shape driven by R-group interactions), quaternary (multiple polypeptide chains assembled into a single functional protein, e.g. hemoglobin). Functions include enzyme catalysis, immune defense, transport, and structural support.
• Nucleic acids: Monomers are nucleotides, made of a pentose sugar, phosphate group, and nitrogenous base (A, T, C, G for DNA; A, U, C, G for RNA). DNA is a double-stranded antiparallel helix with complementary base pairing (A-T, C-G); RNA is single-stranded, and carries genetic information from DNA to ribosomes for protein synthesis.

Worked example: Identify the biomolecule described: it contains carbon, hydrogen, and oxygen only; it has a structural role in plant cell walls; it cannot be digested by humans. [1 mark] Solution: Cellulose [1 mark]. Explanation: Cellulose is a polysaccharide made of beta-glucose subunits, forms rigid cell wall fibers, and humans lack the cellulase enzyme required to break its glycosidic bonds.

4. DNA replication, transcription, translation

These three core processes form the central dogma of molecular biology, describing the flow of genetic information from DNA to RNA to protein:

1. DNA replication: Semi-conservative (each new DNA molecule contains one original parent strand and one new daughter strand). Helicase unwinds the double helix and breaks hydrogen bonds between complementary bases; DNA polymerase adds complementary nucleotides to the 3' end of growing strands, so synthesis only occurs in the 5' → 3' direction. The leading strand is synthesized continuously, while the lagging strand is synthesized as short Okazaki fragments joined by DNA ligase. The Meselson-Stahl experiment (using N-15 and N-14 isotopes) provided definitive proof of semi-conservative replication.
2. Transcription: RNA polymerase binds to a promoter sequence on the template DNA strand, unwinds the double helix, and synthesizes a complementary mRNA strand in the 5' → 3' direction. In eukaryotes, non-coding introns are spliced out of pre-mRNA to form mature mRNA, which exits the nucleus to the cytoplasm.
3. Translation: Mature mRNA binds to a ribosome. tRNA molecules carry specific amino acids to the ribosome, matching their anticodon to complementary mRNA codons. Peptide bonds form between adjacent amino acids to build a polypeptide chain, until a stop codon is reached and the polypeptide is released. The genetic code is degenerate (multiple codons can code for the same amino acid), universal (used by all living organisms), and non-overlapping.

Worked example: A template DNA strand has the sequence 3' TAC GAT CCA 5'. Give the corresponding mRNA sequence and amino acid sequence (use the following codon table: AUG = Met, CUA = Leu, GGU = Gly, UAA = Stop). [2 marks] Solution: mRNA sequence is 5' AUG CUA GGU 3' [1 mark]; amino acid sequence is Met - Leu - Gly [1 mark]. Exam tip: Always confirm if a given DNA strand is the coding strand (5'→3', identical to mRNA except T→U) or template strand (3'→5', complementary to mRNA) before writing your answer.

5. Enzymes and metabolic pathways

Enzymes are globular proteins that act as biological catalysts, lowering the activation energy of chemical reactions without being consumed in the process. Each enzyme has an active site with a 3D shape complementary to a specific substrate, following the induced-fit model (the active site changes shape slightly to bind the substrate more tightly). Factors affecting enzyme activity include:

• Temperature: Activity increases up to an optimum temperature, then drops sharply as high temperatures break hydrogen and R-group bonds, denaturing the enzyme.
• pH: Activity peaks at an optimum pH, as extreme pH changes alter R-group charges and denature the enzyme.
• Substrate concentration: Activity increases with substrate concentration until all active sites are saturated, reaching the maximum reaction rate ($V_{\text{max}}$).

Inhibitors reduce enzyme activity: competitive inhibitors bind to the active site, competing with the substrate, so $V_{\text{max}}$ remains the same but the substrate concentration needed to reach half $V_{\text{max}}$ ($K_m$) increases. Non-competitive inhibitors bind to an allosteric site away from the active site, changing the active site shape so substrates cannot bind, so $V_{\text{max}}$ decreases while $K_m$ stays the same. Metabolic pathways are sequential chains or cycles of enzyme-catalyzed reactions, often regulated by end-product inhibition (feedback inhibition): the final product of the pathway binds to an allosteric site on the first enzyme in the pathway, reducing its activity to prevent overproduction of the end product and conserve cellular resources.

Worked example: A student adds a competitive inhibitor to an enzyme reaction that is running at $V_{\text{max}}$. Explain the immediate effect on reaction rate, and how the rate can be restored to $V_{\text{max}}$. [2 marks] Solution: The reaction rate will decrease immediately, as the inhibitor occupies a portion of the enzyme active sites [1 mark]. Adding excess substrate will outcompete the inhibitor for active sites, restoring the reaction rate to $V_{\text{max}}$ [1 mark].

6. Photosynthesis and respiration overviews

Respiration and photosynthesis are the core energy transformation pathways that support all life on Earth:

1. Cellular respiration: The controlled release of energy from organic compounds to produce ATP, the cell’s universal energy currency.

• Anaerobic respiration: Occurs in the cytoplasm in the absence of oxygen, produces 2 net ATP per glucose molecule. In animals, it produces lactate; in yeast and plants, it produces ethanol and carbon dioxide.
• Aerobic respiration: Occurs in the mitochondria, produces 30-38 net ATP per glucose molecule. Steps: glycolysis (glucose split into pyruvate, 2 ATP), link reaction (pyruvate converted to acetyl CoA, carbon dioxide released), Krebs cycle (acetyl CoA oxidized to produce carbon dioxide, NADH, FADH2, 2 ATP), oxidative phosphorylation (electrons from NADH/FADH2 move through the electron transport chain, creating a proton gradient that drives ATP synthesis via chemiosmosis, oxygen is the final electron acceptor).

2. Photosynthesis: Converts light energy into chemical energy stored in glucose, occurring in chloroplasts of autotrophs.

• Light-dependent reactions: Occur in the thylakoid membrane. Light energy excites chlorophyll electrons, water is split via photolysis to release oxygen, protons, and electrons, and ATP and reduced NADP (NADPH) are produced.
• Light-independent reactions (Calvin cycle): Occur in the stroma. Carbon dioxide is fixed by the enzyme RuBisCO, combined with RuBP to form glycerate-3-phosphate, which is converted to triose phosphate using ATP and NADPH. One out of every 6 triose phosphate molecules is used to build glucose, the rest are used to regenerate RuBP to continue the cycle.

Worked example: Calculate the net ATP produced from the aerobic respiration of 5 glucose molecules, assuming 3 ATP per NADH and 2 ATP per FADH2. [1 mark] Solution: 1 glucose produces 38 ATP, so 5 * 38 = 190 net ATP [1 mark]. Exam tip: Examiners accept 30-38 ATP per glucose depending on the shuttle system used, so state your assumption if asked to calculate total ATP yield.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Describing water properties without linking them to hydrogen bonding. Why students do it: They memorize properties in isolation without understanding their root cause. Correct move: Every time you name a water property (e.g. high latent heat of vaporization), add "due to the large number of hydrogen bonds between water molecules that require high energy to break".
• Wrong move: Confusing DNA coding and template strands for transcription. Why: Questions often do not specify which strand is provided, and students assume all DNA sequences are coding strands. Correct move: First check strand direction: 3'→5' strands are templates, 5'→3' strands are coding. If no direction is given, explicitly state your assumption in your response.
• Wrong move: Mixing up competitive and non-competitive inhibitor effects on $V_{\text{max}}$ and $K_m$. Why: Students memorize graphs without understanding the underlying mechanism. Correct move: Remember competitive inhibitors compete for active sites, so high substrate can outcompete them → same $V_{\text{max}}$, higher $K_m$; non-competitive inhibitors reduce the number of functional enzymes → lower $V_{\text{max}}$, same $K_m$.
• Wrong move: Stating that plants only respire in the dark. Why: Students confuse net gas exchange with individual process activity. Correct move: Respiration occurs 24/7 in all living cells, including plant cells. Net carbon dioxide uptake only occurs when photosynthesis rate exceeds respiration rate during daylight hours.
• Wrong move: Forgetting that stop codons do not code for amino acids. Why: Students list all codons when translating mRNA sequences. **Correct move: Always stop translation when you reach UAA, UAG, or UGA, and do not include an amino acid for these codons.

8. Practice Questions (IB Biology HL Style)

Q1

(a) Explain two properties of water that make it an ideal transport medium for mammalian blood. [3 marks] (b) Outline one role of water as a habitat for aquatic organisms. [2 marks]

Worked Solution

(a) 1 mark per property plus 1 mark for link to blood transport:

1. Solvent properties: Water is polar, so ionic nutrients (e.g. sodium, glucose) and polar molecules (e.g. amino acids, hormones) dissolve easily in blood plasma for transport around the body.
2. Thermal stability: High specific heat capacity allows blood to carry heat around the body without large temperature changes, supporting thermoregulation. (b) Ice is less dense than liquid water, so it floats on the surface of water bodies in cold climates [1 mark], insulating the water below and preventing it from freezing solid, allowing aquatic organisms to survive through winter [1 mark].

Q2

An enzyme-catalyzed reaction has a $V_{\text{max}}$ of $120\mu {\text{mol min}}^{-1}$ and a $K_m$ of $40{\text{ mmol dm}}^{-3}$. (a) Define the term $V_{\text{max}}$. [1 mark] (b) A non-competitive inhibitor is added that reduces the number of functional enzymes by 50%. State the new $V_{\text{max}}$ and $K_m$ of the reaction. [2 marks] (c) Explain why the $K_m$ does or does not change in the presence of this inhibitor. [1 mark]

Worked Solution

(a) $V_{\text{max}}$ is the maximum reaction rate, reached when all available enzyme active sites are saturated with substrate [1 mark]. (b) New $V_{\text{max}}=60\mu {\text{mol min}}^{-1}$ [1 mark], new $K_m=40{\text{ mmol dm}}^{-3}$ [1 mark]. (c) $K_m$ remains the same because the active sites of the remaining functional enzymes still have the same affinity for the substrate [1 mark].

Q3

A point mutation changes the 6th nucleotide of a coding DNA sequence 5' ATG CCT GAA TAA 3' from T to A. (a) Give the mutated mRNA sequence. [1 mark] (b) The original amino acid sequence is Met - Pro - Glu. Use the codon table (CCU/CCC/CCA/CCG = Pro, GAA/GAG = Glu, GUA = Val) to give the mutated amino acid sequence. [1 mark] (c) State the type of mutation described. [1 mark]

Worked Solution

(a) Mutated coding DNA: 5' ATG CCA GAA TAA 3', so mRNA is 5' AUG CCA GAA UAA 3' [1 mark]. (b) Mutated amino acid sequence: Met - Pro - Glu [1 mark, as CCA also codes for proline]. (c) Silent mutation [1 mark, as the mutation does not change the amino acid sequence of the polypeptide].

9. Quick Reference Cheatsheet

10. What's Next

Mastery of molecular biology is a prerequisite for almost every other topic in the IB Biology HL syllabus. It directly connects to Topic 3 and 10 (Genetics), where you will explore how mutations in DNA sequences alter protein function and drive inheritance, Topic 1 and 11 (Cell Biology), where you will apply your knowledge of phospholipid bilayers and membrane transport, and Topic 6 and 12 (Human Physiology), where enzyme function and metabolic pathways are core to understanding digestion, gas exchange, and homeostasis. This content is also essential if you choose to complete an Internal Assessment (IA) focused on enzyme activity, photosynthesis rates, or biomolecule identification. If you have gaps in your understanding of any content in this guide, or want to practice more targeted exam-style questions tailored to your weak areas, you can ask Ollie for support at any time on the homepage. You can also move on to our dedicated IB Biology HL study guides for genetics and cell biology next to build out your full subject knowledge base.