describe the semi-conservative replication of DNA during the S phase of the cell cycle, including: the roles of DNA polymerase and DNA ligase (knowledge of other enzymes in DNA replication in cells and different types of DNA polymerase is not expecte

Cambridge International AS & A‑Level Biology 9700 – Complete Revision Notes

How to Use These Notes

• Each topic is presented with Learning Outcomes (LOs) that map directly to the official syllabus statements.
• At the start of every topic you will find the relevant Assessment Objectives (AO):

  • AO1 – Knowledge and understanding
  • AO2 – Application of knowledge (including quantitative work)
  • AO3 – Analysis, evaluation and experimental design (Paper 3 & Paper 5)

• Key concepts are highlighted in bullet points; diagrams are suggested where visualisation aids learning.
• “Math Box” sections give the short calculations expected for AO2.
• Practical investigation outlines provide ideas for Paper 3 (practical skills) and Paper 5 (planning & evaluation).

Assessment‑Objective Overview

1. Cell Structure

• Learning Outcomes

  • Describe the structural organisation of prokaryotic and eukaryotic cells.
  • Explain the function of major organelles (nucleus, mitochondria, chloroplasts, ER, Golgi, vacuoles, ribosomes, cytoskeleton).
  • Discuss the status of viruses in relation to the cell theory.
  • Apply knowledge of light‑microscopy to calculate magnification and resolution.

• Key Points (AO1)

  • Prokaryotes lack a true nucleus and membrane‑bound organelles; DNA is a circular nucleoid.
  • Eukaryotic cells have compartmentalisation that allows specialised metabolic pathways.
  • Viruses are non‑cellular infectious particles; they contain nucleic acid surrounded by protein (and sometimes lipid) and replicate only inside host cells – a point of debate in the definition of “cell”.

• Microscope Sub‑section (AO2)

  • Magnification = ocular lens × objective lens (e.g., 10× × 40× = 400×).
  • Resolution (minimum distance between two points) ≈ 0.2 µm for light microscopes (λ/2NA).
  • Interpretation of light‑microscope images (e.g., onion epidermis) and electron‑microscope images (e.g., mitochondria cristae).

• Practical (AO3) – Light‑microscopy of onion epidermal cells; sketch, label structures and calculate total magnification.

2. Biomolecules

• Learning Outcomes

  • Identify the four major classes of biomolecules and their monomers.
  • Explain the relationship between structure and function for carbohydrates, lipids, proteins and nucleic acids.

• Key Points (AO1)

  • Polysaccharides are polymers of monosaccharides joined by glycosidic bonds.
  • Phospholipids form bilayers because of their amphipathic nature.
  • Proteins are polymers of amino acids; the sequence of R‑groups determines folding and function.
  • Nucleic acids are polymers of nucleotides; the phosphate‑deoxyribose‑base structure underpins DNA stability and information storage.

• Math Box (AO2) – Using the formula CₙH₂ₙ₊₂Oₙ, calculate the molecular weight of glucose (C₆H₁₂O₆).
  Answer: (6 × 12.01) + (12 × 1.008) + (6 × 16.00) ≈ 180 g mol⁻¹.
• Practical (AO3) – Iodine test for starch; Benedict’s test for reducing sugars.

3. Enzymes

• Learning Outcomes

  • Describe how enzymes lower activation energy and the nature of the enzyme–substrate complex.
  • Interpret enzyme kinetic graphs (Michaelis–Menten, Lineweaver‑Burk).
  • Discuss factors affecting enzyme activity (temperature, pH, inhibitors, substrate concentration).

• Key Points (AO1)

  • Enzymes are biological catalysts; they are not consumed in the reaction.
  • Competitive inhibitors resemble the substrate and bind the active site.

• Math Box (AO2)

  Given the data below, plot a Michaelis–Menten curve and estimate Vmax and Km. Use a Lineweaver‑Burk plot to calculate the values precisely.

  [S] (mM)	Rate (µmol min⁻¹)
  0.5	12
  1.0	20
  2.0	30
  5.0	38
  10.0	40

• Practical (AO3) – Effect of temperature on the rate of catalase activity (hydrogen peroxide breakdown).

4. Cell Membranes & Transport

• Learning Outcomes

  • Explain the fluid‑mosaic model of the plasma membrane.
  • Describe passive (diffusion, osmosis) and active transport mechanisms (pumps, co‑transport, endocytosis, exocytosis).
  • Identify the functional roles of cholesterol, glycolipids and glycoproteins.
  • Outline a typical cell‑signalling cascade.

• Key Points (AO1)

  • Transport proteins are integral membrane proteins that facilitate selective permeability.
  • ATP‑driven Na⁺/K⁺‑ATPase maintains electrochemical gradients.
  • Cholesterol modulates membrane fluidity; glycolipids and glycoproteins are involved in cell‑cell recognition and signalling.
  • Typical signalling pathway: ligand binds receptor → conformational change → second messenger (cAMP, Ca²⁺) → protein kinase cascade → cellular response.

• Math Box (AO2) – Calculate the osmotic pressure (π) of a 0.150 M glucose solution at 25 °C using π = iMRT (i = 1, R = 0.0821 L·atm·mol⁻¹·K⁻¹, T = 298 K).
  Answer: π ≈ 3.68 atm.
• Practical (AO3) – Investigate the effect of solute concentration on the rate of diffusion using dialysis tubing.

5. Cell Division (Mitosis & Meiosis)

• Learning Outcomes

  • Outline the stages of mitosis and describe the changes in chromosome number and arrangement.
  • Contrast mitosis with meiosis I and II, emphasising genetic variation.
  • Explain the role of telomeres, stem cells and the consequences of uncontrolled division.

• Key Points (AO1)

  • During mitosis, sister chromatids separate; the daughter cells are diploid (2n).
  • Meiosis produces four haploid (n) gametes and includes crossing‑over (genetic recombination).
  • Telomeres protect chromosome ends; they shorten with each S‑phase and are lengthened by telomerase in germ‑line and stem cells.
  • Stem cells retain the ability to divide indefinitely and differentiate, providing a source for tissue repair.
  • Loss of cell‑cycle control → uncontrolled division → tumour formation.

• AO2 Example – Predict the chromosome complement of a gamete produced by a plant with 2n = 24.
  Answer: n = 12 chromosomes.
• Practical (AO3) – Prepare and analyse onion root tip squashes to identify mitotic phases.

6. Nucleic Acids & DNA Replication (Semi‑Conservative)

• Learning Outcomes (AO1, AO2, AO3)

  • Define nucleotides and describe the structure of DNA and RNA.
  • Explain the semi‑conservative model of DNA replication.
  • Describe the specific roles of DNA polymerase and DNA ligase.
  • Differentiate leading‑strand from lagging‑strand synthesis.
  • Interpret a replication‑fork diagram and predict the outcome of a mutation that inactivates DNA ligase.
  • Design a simple experiment to demonstrate DNA extraction from plant cells (Paper 5).
  • Calculate the molecular mass of a nucleotide (AO2).
  • Outline RNA processing (capping, poly‑A tail, splicing of introns).

• Key Concepts (AO1)

  • Nucleotide structure – phosphate group, deoxyribose (DNA) or ribose (RNA), nitrogenous base (A, T, C, G; U in RNA).
  • DNA double helix – antiparallel strands, complementary base‑pairing (A–T, C–G) stabilised by hydrogen bonds.
  • Semi‑conservative replication – each daughter DNA molecule contains one parental strand and one newly synthesised strand.

• Step‑by‑Step Process (AO2)

  1. Initiation at the origin of replication

     • Specific DNA sequences (origins) are recognised by initiator proteins.
     • Helicase unwinds the double helix, creating two replication forks.

  2. Stabilisation of single strands

     • Single‑strand binding (SSB) proteins prevent re‑annealing and protect the DNA.

  3. Primer synthesis

     • Primase synthesises short RNA primers (5′→3′) providing a free 3′‑OH for DNA polymerase.

  4. Elongation – DNA polymerase activity

     • DNA polymerase adds deoxyribonucleotides to the 3′‑OH of the primer, synthesising DNA only in the 5′→3′ direction.
     • Proofreading (3′→5′ exonuclease) removes mis‑incorporated bases, reducing the error rate to ≈10⁻⁸ per nucleotide.

  5. Leading‑strand synthesis

     • Continuous synthesis towards the replication fork.
     • Only one RNA primer is required at the origin.

  6. Lagging‑strand synthesis

     • Discontinuous synthesis away from the fork, producing Okazaki fragments (≈100–200 nt in eukaryotes).
     • Each fragment begins with a new RNA primer.

  7. Primer removal & fragment joining

     • RNA primers are removed by RNase H (or the 5′→3′ exonuclease activity of DNA polymerase δ) and replaced with DNA.
     • DNA ligase catalyses the formation of phosphodiester bonds between adjacent Okazaki fragments, sealing nicks to produce a continuous strand.

  8. Termination

     • Replication forks meet; topoisomerase (DNA gyrase) relieves super‑coiling and prevents tangling.

• Roles of the Two Enzymes Highlighted (AO1)

  • DNA polymerase – adds nucleotides, ensures directionality (5′→3′), and proofreads.
  • DNA ligase – joins adjacent DNA fragments by forming phosphodiester bonds; essential for completing the lagging strand and for sealing nicks after repair.

• Leading vs. Lagging Strand Comparison (AO1)

  Feature	Leading Strand	Lagging Strand
  Direction of synthesis relative to fork movement	Continuous, same direction as fork (5′→3′)	Discontinuous, opposite direction (short fragments)
  Primer requirement	One primer at origin	Multiple primers – one per Okazaki fragment
  Enzyme activity after synthesis	Polymerase proceeds without interruption	DNA ligase required to join fragments
  Typical fragment length (eukaryotes)	Whole strand	≈100–200 nt (Okazaki fragments)

• What Happens If DNA Ligase Is Inactive? (AO3)

  • Okazaki fragments remain separate; the lagging strand contains numerous nicks.
  • Unsealed nicks can lead to strand breaks during subsequent replication cycles, causing genomic instability and potentially lethal mutations.

• Illustrative Diagram (suggested)

  • Replication fork showing helicase, SSB, primase, DNA polymerase α/δ/ε, RNA primers, Okazaki fragments, DNA ligase, and topoisomerase.

• Math Box (AO2) – Molecular Mass of a Nucleotide

  Calculate the average mass of a deoxyribonucleotide (dAMP). Approximate atomic masses: C = 12.01, H = 1.008, N = 14.01, O = 16.00, P = 30.97.

  dAMP formula: C₁₀H₁₄N₅O₆P

  Mass ≈ (10 × 12.01) + (14 × 1.008) + (5 × 14.01) + (6 × 16.00) + 30.97 ≈ 331 g mol⁻¹.

• RNA Processing (AO1)

  • Capping – addition of a 7‑methylguanosine cap to the 5′ end protects mRNA from degradation.
  • Poly‑A tail – addition of ~200 adenine residues to the 3′ end stabilises mRNA and aids nuclear export.
  • Splicing – removal of introns and ligation of exons by the spliceosome; exons are the coding sequences retained in mature mRNA.

• Practical Investigation (AO3 – Paper 5)

  Extraction of DNA from strawberries

  1. Homogenise 10 g of strawberries in 50 mL extraction buffer (2 % w/v detergent, 0.5 % NaCl, 10 mM EDTA, pH 8).
  2. Add 20 mL chilled 96 % ethanol slowly while mixing; DNA precipitates as a white, fibrous mass.
  3. Spool the DNA with a glass rod, rinse with 70 % ethanol, and allow to air‑dry.

  Evaluation points: effectiveness of detergent in lysing membranes, role of salt in neutralising phosphate charges, sources of contamination, and how the procedure illustrates the semi‑conservative model (each strand serves as a template for new synthesis).

7. Transport in Plants

• Learning Outcomes

  • Explain the structure of xylem and phloem and the mechanisms of water, mineral and sugar transport.
  • Describe adaptations of xerophytic leaves that reduce water loss.

• Key Points (AO1)

  • Root pressure, capillary action and transpiration pull drive water upward in xylem.
  • Phloem transports photosynthates by a pressure‑flow mechanism (source → sink).
  • Xerophytic adaptations: thick waxy cuticle, sunken stomata, reduced leaf surface area, extensive suberised tissue.

• Case‑Study Box (AO2) – Compare a typical mesophytic leaf with a succulent xerophytic leaf; predict differences in transpiration rates.
• Practical (AO3) – Measure the rate of water uptake in celery stalks placed in coloured dye solutions; discuss the role of transpiration.

8. Transport in Mammals

• Learning Outcomes

  • Describe the structure and function of the human circulatory system.
  • Explain gas exchange in the lungs and tissues.
  • Interpret the oxygen‑dissociation curve and the effects of pH, CO₂ and temperature (Bohr shift, chloride shift).

• Key Points (AO1)

  • Heart chambers, valves and the double‑circulatory system ensure unidirectional flow.
  • Haemoglobin binds O₂ cooperatively; the sigmoid oxygen‑dissociation curve reflects this.
  • Bohr shift – increased CO₂ or decreased pH lowers haemoglobin’s affinity for O₂, facilitating release in active tissues.
  • Chloride shift – Cl⁻ moves into red blood cells as HCO₃⁻ exits, maintaining electro‑neutrality during CO₂ transport.

• Math Box (AO2) – Using the Hill equation, calculate the % saturation of haemoglobin at a PO₂ of 40 mmHg (n ≈ 2.8, P₅₀ ≈ 26 mmHg).
  Answer: % ≈ (PO₂ⁿ / (P₅₀ⁿ + PO₂ⁿ)) × 100 ≈ 73 %.
• Practical (AO3) – Measure heart rate before and after mild exercise; plot heart rate against time and discuss the role of the autonomic nervous system.

9. Nutrition

• Learning Outcomes

  • Explain the digestion and absorption of carbohydrates, proteins and lipids.
  • Describe the role of enzymes, bile salts and brush‑border microvilli.
  • Discuss the consequences of malabsorption disorders.

• Key Points (AO1)

  • Amylase, proteases and lipases act at specific pH ranges in the mouth, stomach and small intestine.
  • Bile salts emulsify fats, increasing surface area for pancreatic lipase.
  • Microvilli increase the absorptive surface area > 600 m² in the adult human small intestine.

• Math Box (AO2) – Calculate the total surface area of the small intestine assuming an average length of 6 m, diameter of 2 cm and villi increase the area by a factor of 10.
  Answer: Approx. 600 m².
• Practical (AO3) – Test the activity of pancreatic amylase on starch solutions at different pH values; plot activity vs. pH.

10. Respiration & Metabolism

• Learning Outcomes

  • Describe aerobic and anaerobic respiration pathways.
  • Explain the role of glycolysis, the citric‑acid cycle and oxidative phosphorylation.
  • Interpret a typical ATP‑yield diagram for glucose metabolism.

• Key Points (AO1)

  • Glycolysis occurs in the cytoplasm, yielding 2 ATP and 2 NADH per glucose.
  • The citric‑acid cycle (mitochondrial matrix) generates 2 ATP, 6 NADH, 2 FADH₂ per glucose.
  • Oxidative phosphorylation (inner mitochondrial membrane) produces ~28–34 ATP via the electron‑transport chain.

• Math Box (AO2) – Total ATP from complete aerobic respiration of one glucose molecule (including substrate‑level phosphorylation).
  Answer: 2 + 2 + 28 ≈ 32 ATP (range 30–38 depending on shuttle system).
• Practical (AO3) – Measure the rate of CO₂ production by yeast in sugar solutions of varying concentrations; discuss the effect of substrate concentration on fermentation rate.

11. Genetics & Evolution

• Learning Outcomes

  • Explain Mendelian inheritance and the concepts of genotype, phenotype and allele.
  • Describe the molecular basis of gene expression (transcription, translation, regulation).
  • Discuss mechanisms of evolution (natural selection, genetic drift, gene flow).

• Key Points (AO1)

  • Dominant and recessive alleles determine phenotype; heterozygotes express the dominant trait.
  • Transcription produces a primary RNA transcript; processing yields mature mRNA.
  • Translation reads codons (3‑nt) on mRNA to assemble a polypeptide chain at the ribosome.
  • Regulation can occur at transcriptional (promoter, enhancers) or translational (miRNA) levels.

• Math Box (AO2) – Using a monohybrid cross (Aa × Aa), calculate the expected phenotypic ratio and the probability of obtaining an aa offspring.
  Answer: Phenotypic ratio 3 : 1; probability aa = ¼.
• Practical (AO3) – Perform a test‑cross using Drosophila flies to determine the genotype of a dominant‑phenotype individual; analyse the offspring ratios.

Glossary of Frequently Used Abbreviations

• AO – Assessment Objective
• LO – Learning Outcome
• S‑phase – Synthesis phase of the cell cycle (DNA replication)
• SSB – Single‑strand binding protein
• Okazaki fragments – Short DNA pieces synthesised on the lagging strand
• R‑group – Side chain of an amino acid
• Bohr shift – Decrease in haemoglobin O₂ affinity caused by lower pH or higher CO₂
• Cl⁻ shift – Exchange of chloride ions for bicarbonate ions in red blood cells

How to Use the “Math Box” and “Practical” Sections

• Math Box – Work through the calculation step‑by‑step, write down units, and check your answer against the provided solution.
• Practical – Follow the procedure, record observations in a table, and use the suggested evaluation points to discuss reliability, sources of error and possible improvements.