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Cambridge International AS & A‑Level Biology (9700) – Revised Syllabus Notes

1. Introduction

This document summarises the core knowledge, practical skills and mathematical techniques required for the Cambridge International AS & A‑Level Biology syllabus (9700). It follows the official ordering of topics (1‑19) and highlights the assessment objectives (AO1–AO3). Each section contains:

• Key points – concise bullet‑point summaries of essential facts (AO1).
• Checklist – a quick‑scan list of the specific learning outcomes the syllabus expects for the topic (helps verify AO1 coverage).
• Worked example – a step‑by‑step quantitative problem (AO2).
• Practical tip / technique – relevant experimental skill (AO3).

2. Cell Structure & Microscopy

2.1 Key Points

• Prokaryotes: no nucleus, circular DNA in nucleoid, no membrane‑bound organelles.
• Eukaryotes: nucleus with linear chromosomes, membrane‑bound organelles (mitochondria, ER, Golgi, lysosomes, chloroplasts in plants).
• Plant cells possess a rigid cell wall (cellulose) and a large central vacuole; animal cells have a flexible plasma membrane.

2.2 Checklist – AS Learning Outcomes

2.3 Microscopy – Types & Uses

2.4 Worked Example – Calculating Magnification

Problem: A light microscope uses an ocular lens of 10 × and an objective of 40 ×. What is the total magnification? If a cell measures 20 µm in the image, what is its actual size?

1. Total magnification = ocular × objective = 10 × × 40 × = 400 ×.
2. Actual size = image size ÷ total magnification = 20 µm ÷ 400 = 0.05 µm (i.e., 50 nm).

2.5 Practical Tip

Use the resolution formula d ≈ 0.61 λ / NA (λ = wavelength of light, NA = numerical aperture) to estimate the smallest resolvable detail. For visible light (λ ≈ 550 nm) and NA = 1.25, d ≈ 0.27 µm.

3. Biological Molecules

3.1 Carbohydrates

• Monosaccharides (glucose, fructose) – C₆H₁₂O₆.
• Disaccharides formed by condensation (loss of H₂O): sucrose, maltose.
• Polysaccharides:

  • Starch = amylose (α‑1,4) + amylopectin (α‑1,4 + α‑1,6).
  • Glycogen – highly branched, rapid glucose release.
  • Cellulose – β‑1,4 linkages, structural component of plant cell walls.

3.2 Lipids

• Triglycerides – glycerol + 3 fatty acids (energy storage).
• Phospholipids – amphipathic; form bilayers; critical for membrane fluidity.
• Steroids – cholesterol (membrane stability) and steroid hormones (signalling).

3.3 Proteins

• 20 standard amino acids; peptide bonds link residues.
• Four levels of structure:

  1. Primary – linear sequence.
  2. Secondary – α‑helix, β‑sheet (hydrogen bonding).
  3. Tertiary – 3‑D folding (hydrophobic interactions, disulfide bridges).
  4. Quaternary – assembly of subunits (e.g., hemoglobin).

• Functions: enzymes, transport, structural, signalling, immune defence.

3.4 Nucleic Acids

• DNA – deoxyribose, bases A, T, G, C; antiparallel double helix.
• RNA – ribose, bases A, U, G, C; usually single‑stranded.

3.5 Water

• Polarity → hydrogen bonding → high specific heat, cohesion, adhesion, surface tension.
• Roles: solvent, medium for transport, temperature regulation, reactant in hydrolysis.

3.6 Checklist – AS Outcomes

3.7 Worked Example – Energy from Glucose

Calculate the theoretical ATP yield from complete aerobic oxidation of one molecule of glucose.

1. Glycolysis: 2 ATP (net) + 2 NADH → 2 × 3 = 6 ATP.
2. Pyruvate oxidation (2 × NADH) → 2 × 3 = 6 ATP.
3. Citric‑acid cycle (6 NADH, 2 FADH₂, 2 GTP):

   • 6 NADH → 6 × 3 = 18 ATP.
   • 2 FADH₂ → 2 × 2 = 4 ATP.
   • 2 GTP ≈ 2 ATP.

4. Total = 2 + 6 + 6 + 18 + 4 + 2 = 38 ATP per glucose molecule.

4. Enzymes

4.1 Mechanism of Action

• Lock‑and‑key vs. induced‑fit models – substrate fits a complementary active site, causing a conformational change.
• Enzyme–substrate complex (ES) lowers activation energy (E_a).

4.2 Factors Affecting Activity

4.3 Michaelis–Menten Kinetics

The relationship between initial velocity (v) and substrate concentration ([S]) is:

\$v = \frac{V{\max}[S]}{Km + [S]}\$

• V_max – maximal rate when all enzyme sites are occupied.
• K_m – substrate concentration at which v = ½ V_max; inversely related to affinity.

4.4 Derivation (AO2)

1. Write the elementary steps: E + S ⇌ ES → E + P.
2. Assume steady‑state for ES (d[ES]/dt ≈ 0).
3. Obtain: K_m = (k_-1 + k₂)/k₁.
4. Substitute into the rate equation to yield the Michaelis–Menten form.

4.5 Worked Example – Determining K_m from Data

Given the following initial rates for an enzyme:

Plot 1/[S] vs 1/v (Lineweaver–Burk). The intercept on the 1/v axis gives 1/V_max ≈ 0.025 min µmol⁻¹ → V_max ≈ 40 µmol min⁻¹ mg⁻¹. The x‑intercept is –1/K_m ≈ –0.10 mM⁻¹ → K_m ≈ 10 mM.

4.6 Practical Tip – Catalase Assay (Paper 5)

Measure the volume of O₂ produced from H₂O₂ at 10 °C, 25 °C and 40 °C. Plot rate vs temperature to locate the optimum and illustrate thermal denaturation.

5. Cell Membranes & Transport

5.1 Fluid‑Mosaic Model

• Phospholipid bilayer with embedded integral and peripheral proteins.
• Cholesterol modulates fluidity; glycoproteins act as receptors.

5.2 Transport Mechanisms

5.3 Quantitative Example – Diffusion Rate

Rate of diffusion (J) ∝ (ΔC × A) / d, where ΔC = concentration difference, A = membrane area, d = thickness.

Given: ΔC = 0.02 mol L⁻¹, A = 1 × 10⁻⁶ m², d = 5 nm. Assuming a proportionality constant k = 1 × 10⁻⁸ m s⁻¹,

J = k × (ΔC × A / d) = 1 × 10⁻⁸ × (0.02 × 1 × 10⁻⁶ / 5 × 10⁻⁹) ≈ 4 × 10⁻⁸ mol s⁻¹.

5.4 Practical Skill – Plasmolysis (Paper 3)

1. Place an onion epidermal strip in 0.5 M sucrose.
2. Observe cell shrinkage under a light microscope.
3. Record the change in cell length; calculate % plasmolysis.

5.5 Checklist – AS Outcomes

6. The Cell Cycle & Mitosis

6.1 Phases of the Cell Cycle

• Interphase: G₁ (growth), S (DNA synthesis), G₂ (pre‑mitosis).
• Mitosis: Prophase, Metaphase, Anaphase, Telophase.
• Cytokinesis: Division of cytoplasm – cleavage furrow (animal) or cell plate (plant).

6.2 Detailed Features (AO2)

6.3 Telomeres & Chromosome Stability (A‑Level)

• Telomeres are repetitive DNA sequences at chromosome ends that protect against degradation.
• Telomerase adds repeats in germ cells, stem cells and many cancer cells.

6.4 Worked Example – Calculating Mitotic Index

In a sample of 500 root‑tip cells, 25 are observed in mitosis. Mitotic index = (25 / 500) × 100 = 5 %. This value can be compared between healthy and chemically‑treated tissues.

6.5 Practical Skill – Onion Root‑Tip Squash (Paper 3)

1. Harvest root tip (≈1 cm), treat with 1 % colchicine for 2 h.
2. Fix in Carnoy’s solution, hydrolyse in 1 N HCl (60 °C, 10 min).
3. Stain with aceto‑orcein, mount and observe under 400 × magnification.
4. Record the proportion of cells in each mitotic stage.

6.6 Checklist – AS Outcomes

7. DNA, RNA & Protein Synthesis

7.1 DNA Structure & Replication

• Double helix: antiparallel strands, complementary base‑pairing (A‑T, G‑C).
• Replication is semi‑conservative: each daughter DNA contains one parental strand.

Steps of Replication (AO2)

1. Unwinding by helicase; single‑strand binding proteins stabilise.
2. RNA primer synthesis by primase.
3. Leading‑strand synthesis – continuous (DNA polymerase III).
4. Lagging‑strand synthesis – discontinuous (Okazaki fragments).
5. Removal of RNA primers (DNA polymerase I) and replacement with DNA.
6. Ligation of fragments (DNA ligase).

7.2 Transcription (Nucleus)

• RNA polymerase binds promoter, unwinds DNA, synthesises complementary RNA (U replaces T).
• RNA processing in eukaryotes:

  • 5′ cap addition.
  • Splicing – removal of introns.
  • Poly‑A tail addition.

7.3 Translation (Cytoplasm)

1. mRNA attaches to the small ribosomal subunit.
2. Initiator tRNA (Met) pairs with the start codon (AUG).
3. Large subunit joins; peptide bond formation by peptidyl‑transferase.
4. Elongation – successive tRNAs bring amino acids matching codons.
5. Termination – stop codon (UAA, UAG, UGA) recognised by release factors.
6. Polypeptide folds, may undergo post‑translational modifications.

7.4 The Genetic Code (Degeneracy)

7.5 Worked Example – Predicting a Polypeptide

mRNA sequence: 5'‑AUG‑GCC‑UAA‑3'. Translate:

1. AUG → Met (start).
2. GCC → Ala.
3. UAA → Stop.

Resulting peptide: Met‑Ala.

7.6 Practical Skill – DNA Extraction (Paper 5)

Extract DNA from ripe strawberries using detergent, salt and ethanol precipitation. Visualise the white DNA thread and discuss the role of each reagent.

7.7 Checklist – AS Outcomes

8. Plant Transport

8.1 Xylem – Water & Mineral Transport

• Root pressure: positive pressure generated by active ion uptake in roots; can push water upward (e.g., exudation in cut wheat stems).
• Cohesion‑tension theory: transpiration creates a negative pressure (tension) that pulls water upward; cohesion between water molecules and adhesion to vessel walls maintain a continuous column.
• Structure: tracheids (long, tapered, pits) and vessel elements (short, perforation plates) – both lignified.

8.2 Phloem – Photoassimilate Transport

• Pressure‑flow hypothesis: loading of sucrose at source raises osmotic pressure, water follows, generating bulk flow toward sink where sucrose is unloaded.
• Sieve‑tube elements (enucleate) + companion cells (metabolically active).

8.3 Quantitative Example – Transpiration Rate

Given a leaf area of 30 cm², a water loss of 0.45 g h⁻¹, calculate the transpiration rate in mmol m⁻² s⁻¹.

1. Convert mass to moles: 0.45 g ÷ 18 g mol⁻¹ = 0.025 mol h⁻¹.
2. Convert hours to seconds: 0.025 mol ÷ 3600 s ≈ 6.9 × 10⁻⁶ mol s⁻¹.
3. Area in m²: 30 cm² = 3 × 10⁻³ m².
4. Rate = (6.9 × 10⁻⁶ mol s⁻¹) ÷ 3 × 10⁻³ m² ≈ 2.3 × 10⁻³ mol m⁻² s⁻¹ = 2.3 mmol m⁻² s⁻¹.

8.4 Practical Demonstration – Celery Dye Transport

1. Place a celery stalk in coloured potassium nitrate solution.
2. After 24 h, cut transverse sections and observe the upward movement of dye through xylem.
3. Discuss the role of transpiration pull.

8.5 Checklist – AS Outcomes

9. Animal Transport & Gas Exchange

9.1 Circulatory System

• Heart chambers: 2 (fish), 3 (amphibians), 4 (mammals, birds).
• Blood vessels: arteries (high pressure), veins (low pressure, valves), capillaries (site of exchange).
• Plasma: water, ions, proteins (albumin, globulins, fibrinogen).
• Red blood cells contain haemoglobin – each haemoglobin molecule binds up to 4 O₂ molecules.

9.2 Gas Exchange

• Diffusion across respiratory surfaces (alveoli, gills, insect tracheae).
• Factors influencing rate: surface area (A), thickness (d), partial pressure difference (Δp), diffusion coefficient (D). Fick’s law: Rate = (D × A × Δp) / d.

9.3 Worked Example – Respiratory Quotient (RQ)

During a 5‑min exercise test a subject consumes 0.30 L O₂ and produces 0.24 L CO₂. RQ = VCO₂ / VO₂ = 0.24 / 0.30 = 0.80, indicating predominant carbohydrate metabolism (RQ ≈ 1) with a contribution from fat oxidation (RQ ≈ 0.7).

9.4 Practical Skill – Measuring CO₂ Diffusion (Paper 3)

1. Fill a sealed syringe with water and a known volume of CO₂.
2. Insert a fine glass tube containing a known mass of sodium bicarbonate.
3. Measure the change in gas volume over time at 20 °C and 30 °C.
4. Calculate the diffusion coefficient using Fick’s law.

9.5 Checklist – AS Outcomes

10. Respiration

10.1 Overview

• Aerobic respiration: C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + ≈ 38 ATP.
• Anaerobic pathways:

  • Lactic‑acid fermentation (muscle): glucose → 2 lactate + 2 ATP.
  • Alcoholic fermentation (yeast): glucose → 2 ethanol + 2 CO₂ + 2 ATP.

10.2 Glycolysis (Cytosol)

1. Glucose → glucose‑6‑phosphate (hexokinase, ATP used).
2. Fructose‑6‑phosphate → fructose‑1,6