Skip to content

Biology - Cell Biology and Biochemistry

  1. All living organisms are composed of one or more cells
  2. The cell is the basic unit of structure and function in all living organisms
  3. All cells arise from pre-existing cells
FeatureProkaryotic CellsEukaryotic Cells
NucleusNo true nucleus (nucleoid region)True nucleus with nuclear envelope
Membrane-bound organellesAbsentPresent
DNACircular, nakedLinear, associated with histones
Ribosomes70S (smaller)80S (larger)
Cell size1-5 micrometres10-100 micrometres
Cell wallPresent (peptidoglycan in bacteria)Present in plants (cellulose), absent in animals
ExamplesBacteria, ArchaeaAnimals, Plants, Fungi, Protists
FeaturePlant CellsAnimal Cells
Cell wallPresent (cellulose)Absent
ChloroplastsPresentAbsent
Large central vacuolePresentSmall, temporary vacuoles
CentriolesAbsentPresent
PlasmodesmataPresentAbsent
ShapeFixed (rectangular)Variable (irregular)
Stored carbohydrateStarchGlycogen

The nucleus is the control centre of the cell. It contains:

  • Nuclear envelope: Double membrane with nuclear pores that control the movement of substances in and out
  • Nucleolus: Dark-staining region where ribosomal RNA (rRNA) is synthesised
  • Chromatin: DNA wrapped around histone proteins; condenses into chromosomes during cell division

Functions:

  • Controls cell activities through gene expression
  • Stores genetic information (DNA)
  • Site of DNA replication and transcription

Mitochondria are the sites of aerobic respiration, producing ATP through the Krebs cycle and Oxidative phosphorylation.

Structure:

  • Outer membrane: Permeable to small molecules
  • Inner membrane: Folded into cristae to increase surface area; site of the electron transport chain
  • Matrix: Contains enzymes for the Krebs cycle, mitochondrial DNA, and ribosomes

C6H12O6+6O26CO2+6H2O+ATP\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2 \to 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} + \mathrm{ATP}

Rough ER (RER):

  • Studded with ribosomes
  • Involved in protein synthesis and transport
  • Proteins enter the ER lumen for folding and modification

Smooth ER (SER):

  • No ribosomes
  • Involved in lipid synthesis, detoxification, and calcium storage

A stack of flattened membrane-bound sacs (cisternae) that modifies, sorts, and packages proteins and Lipids.

Functions:

  • Modifies proteins (e.g., adding carbohydrate groups to form glycoproteins)
  • Packages proteins into vesicles for transport
  • Forms lysosomes (in animal cells)

Ribosomes are the sites of protein synthesis (translation).

  • Free ribosomes: Float in the cytoplasm; synthesise proteins for use within the cell
  • Bound ribosomes: Attached to the RER; synthesise proteins for secretion or for the cell membrane

Membrane-bound vesicles containing digestive enzymes (hydrolytic enzymes). Functions include:

  • Breaking down worn-out organelles (autophagy)
  • Digesting material engulfed by phagocytosis
  • Releasing enzymes outside the cell (extracellular digestion)

The site of photosynthesis.

Structure:

  • Double membrane: Outer and inner membrane
  • Thylakoids: Flattened sacs containing chlorophyll; site of the light-dependent reactions
  • Grana: Stacks of thylakoids
  • Stroma: Fluid-filled space; site of the light-independent reactions (Calvin cycle)

6CO2+6H2OlightC6H12O6+6O26\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{light}} \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2

The cell membrane (plasma membrane) is a phospholipid bilayer with embedded proteins.

Fluid Mosaic Model:

  • Fluid: Individual phospholipid molecules can move laterally within the layer
  • Mosaic: Various proteins are embedded in the bilayer, creating a pattern

Components:

  • Phospholipids: form the basic bilayer structure
  • Intrinsic (integral) proteins: span the entire membrane; involved in transport and signalling
  • Extrinsic (peripheral) proteins: attached to the surface; involved in cell recognition and enzymatic activity
  • Cholesterol: regulates membrane fluidity
  • Glycolipids and glycoproteins: involved in cell recognition

Movement of molecules from a region of higher concentration to a region of lower concentration, Directly through the phospholipid bilayer.

  • Only for small, non-polar molecules (e.g., O2\mathrm{O}_2, CO2\mathrm{CO}_2Lipid-soluble substances)
  • Net movement stops at equilibrium (dynamic equilibrium)
  • Rate depends on: concentration gradient, temperature, surface area, distance

Movement of molecules down their concentration gradient through transport proteins.

  • Channel proteins: Form hydrophilic pores for specific ions (e.g., Na+\mathrm{Na}^+ K+\mathrm{K}^+, Ca2+\mathrm{Ca}^{2+})
  • Carrier proteins: Bind to specific molecules, undergo conformational change, and release on the other side
  • Used for larger or polar molecules (e.g., glucose, amino acids, ions)

The movement of water molecules from a region of higher water potential to a region of lower water Potential, through a selectively permeable membrane.

Water potential (Ψ\Psi) is measured in kilopascals (kPa). Pure water has Ψ=0kPa\Psi = 0 \mathrm{ kPa}. Adding solutes decreases water potential (makes it more negative).

Osmosis in animal cells:

SolutionWater PotentialEffect on Cell
HypotonicHigher than cellCell swells and may burst (lysis)
IsotonicEqual to cellNo net movement
HypertonicLower than cellCell shrinks (crenation)

Osmosis in plant cells:

SolutionEffect on Cell
HypotonicCell becomes turgid (firm); useful for plant support
IsotonicNo net movement
HypertonicCytoplasm and vacuole shrink; cell becomes plasmolysed

Movement of molecules against their concentration gradient (from low to high concentration), Requiring ATP and carrier proteins.

  • Specific to certain molecules (e.g., Na+/K+\mathrm{Na}^+/\mathrm{K}^+ pump, absorption of minerals by root hair cells)
  • Can become saturated at high concentrations (carrier proteins are limiting)
  • Inhibited by metabolic poisons (e.g., cyanide) that block ATP production

The cell membrane engulfs material to bring it into the cell.

  • Phagocytosis: “Cell eating” — engulfing solid particles (e.g., white blood cells engulfing bacteria)
  • Pinocytosis: “Cell drinking” — engulfing liquid droplets

Vesicles fuse with the cell membrane to release contents outside the cell (e.g., secretion of Hormones, neurotransmitters).


General formula: Cx(H2O)y\mathrm{C}_x(\mathrm{H}_2\mathrm{O})_y

Monosaccharides:

  • Glucose (C6H12O6\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6): primary energy source
  • Fructose: found in fruits
  • Galactose: component of milk sugar
  • Ribose and deoxyribose: in RNA and DNA respectively

Disaccharides (formed by condensation, broken by hydrolysis):

DisaccharideComponent Monosaccharides
MaltoseGlucose + Glucose
SucroseGlucose + Fructose
LactoseGlucose + Galactose

Polysaccharides:

  • Starch: Energy storage in plants (amylose: helical; amylopectin: branched)
  • Glycogen: Energy storage in animals (highly branched)
  • Cellulose: Structural in plants (beta-glucose, straight chains, hydrogen bonds)
  • Chitin: Structural in arthropod exoskeletons and fungal cell walls

Tests for carbohydrates:

  • Benedict”s test: Reducing sugars (e.g., glucose) produce a brick-red precipitate when heated with Benedict’s reagent
  • Iodine test: Starch turns blue-black

Proteins are polymers of amino acids.

Amino acid structure:

Each amino acid has an amino group (NH2\mathrm{NH}_2), a carboxyl group (COOH\mathrm{COOH}), a hydrogen Atom, and an R group (side chain), all bonded to a central carbon atom.

Peptide bond formation:

Amino acids join by condensation reactions, forming peptide bonds (CONH\mathrm{-CO-NH-}) and releasing Water.

  • Dipeptide: 2 amino acids
  • Polypeptide: many amino acids

Protein structure levels:

LevelDescriptionBonds
PrimarySequence of amino acidsPeptide bonds
SecondaryAlpha-helix or beta-pleated sheetHydrogen bonds
Tertiary3D folding of the polypeptideH-bonds, ionic bonds, disulphide bridges, hydrophobic interactions
QuaternaryMultiple polypeptide chainsSame as tertiary + more

Tests for proteins:

  • Biuret test: Add Biuret reagent (NaOH + CuSO4_4); violet/purple colour indicates protein

Lipids are organic molecules that are insoluble in water but soluble in organic solvents.

Triglycerides:

  • Formed from one glycerol + three fatty acids
  • Joined by ester bonds (condensation reaction)
  • Energy storage molecules
  • Contain more energy per gram than carbohydrates

Phospholipids:

  • Similar to triglycerides but one fatty acid is replaced by a phosphate group
  • Amphipathic: hydrophilic head (phosphate) and hydrophobic tail (fatty acids)
  • Form the cell membrane bilayer

Saturated vs unsaturated fatty acids:

  • Saturated: no double bonds (C-C), solid at room temperature (animal fats)
  • Unsaturated: one or more double bonds (C=C), liquid at room temperature (plant oils)

Test for lipids:

  • Ethanol emulsion test: Dissolve in ethanol, pour into water; cloudy white emulsion indicates lipids

DNA (Deoxyribonucleic Acid):

  • Double-stranded helix
  • Sugar: deoxyribose
  • Bases: adenine (A), thymine (T), guanine (G), cytosine (C)
  • Base pairing: A-T (2 H-bonds), G-C (3 H-bonds)
  • Stores genetic information

RNA (Ribonucleic Acid):

  • single-stranded
  • Sugar: ribose
  • Bases: A, U (uracil replaces thymine), G, C
  • Types: mRNA (messenger), tRNA (transfer), rRNA (ribosomal)

Enzymes are biological catalysts produced by living cells. They are globular proteins that speed up Biochemical reactions without being consumed.

  • Specific: each enzyme catalyses one particular reaction (or a small group of reactions)
  • Not used up: can be reused many times
  • Work in small amounts
  • Affected by temperature and pH
  • Protein in nature: denatured by extreme conditions

The substrate (reactant) fits into the active site of the enzyme like a key fits into a lock. The Active site has a specific complementary shape to the substrate.

A more refined model: the active site changes shape slightly when the substrate binds, improving the Fit. This lowers the activation energy of the reaction.

Temperature:

  • Rate increases with temperature (up to the optimum)
  • Beyond the optimum, the enzyme denatures (active site changes shape irreversibly)
  • Human enzymes: optimum around 37C37^\circ\mathrm{C}

pH:

  • Each enzyme has an optimum pH
  • Extreme pH causes denaturation by disrupting ionic and hydrogen bonds
  • Pepsin: optimum pH 2 (stomach)
  • Trypsin: optimum pH 8 (small intestine)

Substrate concentration:

  • Increasing substrate concentration increases the rate (up to a point)
  • At saturation, all active sites are occupied; rate plateaus (Vmax)

Enzyme concentration:

  • Increasing enzyme concentration increases the rate (provided substrate is not limiting)

An enzyme has an optimum temperature of 40C40^\circ\mathrm{C}. At 20C20^\circ\mathrm{C}The reaction Rate is 0.30.3 units/s. At 40C40^\circ\mathrm{C}The rate is 1.21.2 units/s. Calculate the Q10 (temperature coefficient).

Q10=Rateat(T+10)RateatTQ_{10} = \frac{\mathrm{Rate at }(T + 10)}{\mathrm{Rate at } T}

Between 20C20^\circ\mathrm{C} and 30C30^\circ\mathrm{C}: Q10Q_{10} might be approximately 2 (typical For biological reactions). Without the 30C30^\circ\mathrm{C} data, we can estimate the overall Effect:

From 20C20^\circ\mathrm{C} to 40C40^\circ\mathrm{C} (a 20C20^\circ\mathrm{C} increase):

Rateincreasefactor=1.20.3=4\mathrm{Rate increase factor} = \frac{1.2}{0.3} = 4

This is consistent with Q102Q_{10} \approx 2 (since 22=42^2 = 4).

Competitive inhibition:

  • Inhibitor has a similar shape to the substrate
  • Competes with the substrate for the active site
  • Can be overcome by increasing substrate concentration
  • Example: malonate inhibiting succinate dehydrogenase

Non-competitive inhibition:

  • Inhibitor binds to a site other than the active site (allosteric site)
  • Changes the shape of the active site
  • Cannot be overcome by increasing substrate concentration
  • Example: heavy metal ions (lead, mercury)

  1. Glucose (6C) is phosphorylated (uses 2 ATP)
  2. Glucose is split into two molecules of triose phosphate (3C)
  3. Triose phosphate is oxidised and dehydrogenated (produces 2 NADH)
  4. Net production: 2 ATP, 2 NADH, 2 pyruvate (3C)
  • Pyruvate (3C) is decarboxylated and dehydrogenated
  • Forms acetyl CoA (2C) + CO2\mathrm{CO}_2 + NADH

For each glucose molecule (two turns of the cycle):

  • 2 CO2\mathrm{CO}_2 released
  • 3 NADH produced per turn (6 total)
  • 1 FADH2_2 produced per turn (2 total)
  • 1 ATP produced per turn (2 total)
  • Regenerates oxaloacetate (4C)

Oxidative Phosphorylation (Inner Mitochondrial Membrane)

Section titled “Oxidative Phosphorylation (Inner Mitochondrial Membrane)”
  • NADH and FADH2_2 donate electrons to the electron transport chain
  • Energy released pumps protons across the inner membrane
  • Protons flow back through ATP synthase, producing ATP
  • Oxygen is the final electron acceptor, forming water
StageATP (net)
Glycolysis2
Krebs cycle2
Oxidative phosphorylation (from NADH)28
Oxidative phosphorylation (from FADH2_2)4
Totalapproximately 36-38

Light-Dependent Reactions (Thylakoid Membrane)

Section titled “Light-Dependent Reactions (Thylakoid Membrane)”
  1. Light energy is absorbed by chlorophyll in Photosystem II
  2. Water is split (photolysis): H2O2H++2e+12O2\mathrm{H}_2\mathrm{O} \to 2\mathrm{H}^+ + 2e^- + \frac{1}{2}\mathrm{O}_2
  3. Electrons pass through the electron transport chain, generating ATP
  4. Light is absorbed by Photosystem I; electrons are re-energised
  5. Electrons reduce NADP+^+ to NADPH

Products: ATP, NADPH, O2\mathrm{O}_2

Light-Independent Reactions / Calvin Cycle (Stroma)

Section titled “Light-Independent Reactions / Calvin Cycle (Stroma)”
  1. CO2\mathrm{CO}_2 is fixed by ribulose bisphosphate (RuBP, 5C) using the enzyme RuBisCO
  2. Forms an unstable 6C compound that splits into two molecules of glycerate-3-phosphate (GP, 3C)
  3. GP is reduced to triose phosphate (TP, 3C) using ATP and NADPH
  4. Some TP is used to make glucose and other organic compounds
  5. Most TP is used to regenerate RuBP (uses ATP)

For every 3 CO2\mathrm{CO}_2 molecules fixed: 1 molecule of triose phosphate (3C) is produced. It Takes 6 CO2\mathrm{CO}_2 molecules to produce 1 molecule of glucose (6C).

FactorEffect
Light intensityIncreases rate up to a plateau (light saturation point)
CO2\mathrm{CO}_2 concentrationIncreases rate up to a plateau
TemperatureIncreases rate up to optimum, then decreases (enzyme denaturation)

Life’s building blocks: Cells are like factories — the nucleus is the management office, mitochondria are power plants, and ribosomes are assembly lines. Biochemical tests identify which molecules are present.

Why it matters: From diagnosing diseases to understanding genetics, cell biology is the foundation of modern medicine and biotechnology.

The key insight: Structure determines function — the shape of a protein determines what it does, and mutations that change shape can cause disease.

flowchart TD
A[1_Cell Biology And Biochemistry] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]
TopicKey ConceptLocation
Prokaryote vs EukaryoteNucleus, organelles, ribosome sizeAll cells
Membrane transportDiffusion, osmosis, active transportCell membrane
CarbohydratesMonosaccharides, disaccharides, polysaccharidesAll organisms
ProteinsAmino acids, peptide bonds, 4 levels of structureAll organisms
EnzymesLock and key, induced fit, denaturationAll organisms
RespirationGlycolysis, Krebs, oxidative phosphorylationMitochondria
PhotosynthesisLight-dependent, Calvin cycleChloroplasts

  • Always distinguish between prokaryotic and eukaryotic cells in comparison questions.
  • For membrane transport, state whether energy is required and the direction of movement relative to the concentration gradient.
  • In enzyme questions, always mention the effect on the active site when discussing denaturation.
  • For photosynthesis limiting factor graphs, identify which factor is limiting at each point on the curve.
  • When describing biochemical pathways (respiration, photosynthesis), name the specific location of each stage.

Explain how the fluid mosaic model of the cell membrane accounts for the following properties: (a) selective permeability, (b) ability of cells to communicate, (c) flexibility for phagocytosis.

If you get this wrong, revise: Cell Organelles — Cell Membrane

Solution

(a) Selective permeability: The phospholipid bilayer allows small, non-polar molecules (O2_2CO2_2) to diffuse through freely, while blocking large or charged molecules. Channel and carrier proteins provide selective pathways for specific ions and polar molecules (e.g., glucose via carrier proteins, Na+^+ via channel proteins). Cholesterol modulates fluidity, maintaining permeability at different temperatures.

(b) Cell communication: Glycoproteins and glycolipids on the cell surface act as recognition markers (e.g., MHC proteins for immune recognition, hormone receptors). Extrinsic proteins on the outer surface can bind signalling molecules, triggering intracellular responses via signal transduction.

(c) Flexibility: The phospholipid molecules can move laterally within the bilayer (fluid nature), allowing the membrane to bend and change shape. This enables vesicle formation during endocytosis (phagocytosis) and exocytosis, and allows cells to change shape (e.g., red blood cells squeezing through capillaries).

Exam-Style Practice Questions

Question 1: Describe the structure of the cell membrane and explain how its structure relates to Its function.

The cell membrane is a phospholipid bilayer with embedded proteins (fluid mosaic model). The Hydrophobic tails face inward and hydrophilic heads face outward, making the membrane a barrier to Water-soluble substances. Channel and carrier proteins allow selective transport. Cholesterol Maintains fluidity. Glycoproteins on the surface enable cell recognition. The fluid nature allows Vesicle formation (endocytosis/exocytosis).

Question 2: Explain the effect of increasing temperature on enzyme activity from 0C0^\circ\mathrm{C} to 60C60^\circ\mathrm{C}.

From 0C0^\circ\mathrm{C} to the optimum (approximately 3740C37-40^\circ\mathrm{C}): increasing Temperature increases kinetic energy, leading to more frequent collisions between enzyme and Substrate molecules, increasing the rate of reaction. Beyond the optimum temperature, the increased Kinetic energy breaks the bonds maintaining the tertiary structure of the enzyme, causing the active Site to change shape (denaturation). The substrate can no longer fit, and the reaction rate Decreases sharply.

Question 3: Compare and contrast aerobic and anaerobic respiration.

Similarities: Both begin with glycolysis; both produce ATP.

Differences: Aerobic respiration requires oxygen and occurs in the mitochondria, producing Approximately 36-38 ATP per glucose with CO2\mathrm{CO}_2 and H2O\mathrm{H}_2\mathrm{O} as By-products. Anaerobic respiration occurs without oxygen, only in the cytoplasm, producing 2 ATP per Glucose. In animals, it produces lactate; in yeast, it produces ethanol and CO2\mathrm{CO}_2.


Mitosis is the process of cell division that produces two genetically identical daughter cells, each With the same number of chromosomes as the parent cell.

Functions of mitosis:

  • Growth: increasing the number of cells in an organism
  • Repair: replacing damaged or dead cells
  • Asexual reproduction: in some organisms
StageKey Events
ProphaseChromatin condenses into chromosomes; nucleolus disappears; centrioles move to poles; spindle fibres form; nuclear membrane begins to break down
MetaphaseChromosomes line up at the equator (metaphase plate); spindle fibres attach to centromeres
AnaphaseSister chromatids separate at the centromere; spindle fibres pull chromatids to opposite poles; chromosomes have a V-shape
TelophaseChromosomes arrive at poles and decondense; nuclear membrane reforms; nucleolus reappears; spindle fibres break down
CytokinesisCytoplasm divides; in animal cells, a cleavage furrow forms; in plant cells, a cell plate forms

A cell in the G2 phase of the cell cycle has 46 chromosomes. How many chromosomes and chromatids Will be present in each daughter cell after mitosis?

After mitosis: each daughter cell has 46 chromosomes (each consisting of one chromatid).

During S phase (before mitosis), each chromosome replicates to form two sister chromatids. So at the Start of mitosis, there are 46 chromosomes (92 chromatids). After anaphase, the chromatids separate, And each daughter cell receives 46 single-chromatid chromosomes.

Meiosis is a type of cell division that produces four genetically different daughter cells, each With half the number of chromosomes of the parent cell. It is essential for sexual reproduction.

Key differences from mitosis:

FeatureMitosisMeiosis
Number of divisionsOneTwo
Number of daughter cellsTwoFour
Chromosome numberSame as parent (diploid)Half of parent (haploid)
Genetic variationIdentical daughter cellsDifferent daughter cells
FunctionGrowth, repair, asexual reproductionProduction of gametes (sex cells)

Sources of genetic variation in meiosis:

  1. Crossing over: During prophase I, homologous chromosomes exchange segments of DNA. This creates new combinations of alleles on the same chromosome.
  2. Independent assortment: During metaphase I, homologous pairs line up randomly at the equator. Different combinations of maternal and paternal chromosomes are distributed to daughter cells.
  3. Random fertilisation: Any sperm can fertilise any egg, further increasing genetic diversity.

Meiosis I (reductional division):

  • Prophase I: Homologous chromosomes pair up (synapsis); crossing over occurs
  • Metaphase I: Homologous pairs line up at the equator
  • Anaphase I: Homologous chromosomes separate (sister chromatids remain together)
  • Telophase I: Two cells form, each with half the chromosome number (but each chromosome still has two chromatids)

Meiosis II (equational division):

  • Prophase II: Chromosomes condense again
  • Metaphase II: Chromosomes line up singly at the equator
  • Anaphase II: Sister chromatids separate
  • Telophase II: Four haploid daughter cells form

The cell cycle describes the sequence of events from one cell division to the next.

PhaseDescriptionDuration (typical)
G1 (Gap 1)Cell growth, normal metabolism, organelle duplicationVariable
S (Synthesis)DNA replication6-8 hours
G2 (Gap 2)Preparation for mitosis, protein synthesis2-4 hours
M (Mitosis)Cell division1-2 hours
CytokinesisCytoplasmic divisionOverlaps with M phase

G1, S, and G2 together are called interphase, which accounts for approximately 90% of the cell Cycle.

The cell cycle is controlled by checkpoints:

  • G1 checkpoint: Checks if the cell is large enough and DNA is undamaged before entering S phase
  • G2 checkpoint: Checks if DNA has been replicated correctly before entering M phase
  • Spindle checkpoint (M checkpoint): Checks if all chromosomes are properly attached to spindle fibres before anaphase

Cancer is a disease caused by uncontrolled cell division, resulting in the formation of tumours.

  • Benign tumours: Grow slowly, remain localised, do not spread
  • Malignant tumours: Grow rapidly, invade surrounding tissues, can spread to other parts of the body (metastasis)

Causes of cancer:

  • Mutations in proto-oncogenes (become oncogenes, promoting cell division)
  • Mutations in tumour suppressor genes (losing their inhibitory function)
  • Exposure to carcinogens: UV radiation, tobacco smoke, certain chemicals, ionising radiation
  • Some viruses (e.g., HPV, hepatitis B and C)

DNA replication is semi-conservative: each new DNA molecule consists of one original strand and one Newly synthesised strand.

  1. Helicase unwinds and separates the double helix by breaking hydrogen bonds between complementary bases
  2. DNA polymerase adds complementary nucleotides to each template strand, following the base pairing rules (A-T, C-G)
  3. Leading strand: Synthesised continuously in the 5-prime to 3-prime direction
  4. Lagging strand: Synthesised in short fragments (Okazaki fragments) that are later joined by DNA ligase
  5. Each new DNA molecule contains one original strand and one new strand

DNA polymerase has a …/1-number-and-algebra/3_proof-and-logicreading function. If an incorrect nucleotide is added, it is removed and Replaced. This gives an error rate of approximately 1 in 10910^9 base pairs.


Transcription is the process of copying the genetic information from DNA to messenger RNA (mRNA).

  1. RNA polymerase binds to the promoter region on the DNA template strand
  2. The DNA double helix unwinds locally
  3. RNA polymerase synthesises a complementary mRNA strand using the DNA template (A pairs with U, T pairs with A, C pairs with G, G pairs with C)
  4. The mRNA molecule is released when RNA polymerase reaches the terminator region
  5. In eukaryotes, the pre-mRNA is processed:
  • A cap is added to the 5-prime end
  • A poly-A tail is added to the 3-prime end
  • Introns (non-coding regions) are removed by splicing
  • Exons (coding regions) are joined together

Translation is the process of synthesising a polypeptide chain from the mRNA template.

  1. The mRNA binds to a ribosome
  2. The ribosome reads the mRNA in codons (groups of three bases)
  3. Transfer RNA (tRNA) molecules bring specific amino acids to the ribosome
  4. Each tRNA has an anticodon that is complementary to the codon on the mRNA
  5. Peptide bonds form between adjacent amino acids (catalysed by peptidyl transferase in the ribosome)
  6. Translation stops when a stop codon (UAA, UAG, UGA) is reached
  7. The polypeptide chain is released and folds into its functional 3D shape
  • The genetic code is degenerate: most amino acids are coded for by more than one codon
  • The genetic code is universal: the same codons code for the same amino acids in nearly all organisms
  • The genetic code is non-overlapping: each base is part of only one codon

More Exam-Style Problems

Question 4: Describe the role of the Golgi apparatus in protein processing and transport.

The Golgi apparatus receives proteins from the rough ER in transport vesicles. It modifies these Proteins by adding carbohydrate groups (glycosylation) to form glycoproteins. It sorts proteins and Packages them into secretory vesicles for export from the cell (exocytosis), or into lysosomes. The Golgi acts as a processing and distribution centre, ensuring proteins reach their correct Destinations.

Question 5: Explain how the structure of the mitochondrion is adapted to its function in aerobic Respiration.

The mitochondrion has a double membrane. The inner membrane is folded into cristae, providing a Large surface area for the electron transport chain and ATP synthase. The matrix contains enzymes For the Krebs cycle and its own DNA and ribosomes, allowing it to produce some of its own proteins. The small intermembrane space allows the accumulation of protons for the chemiosmotic gradient. These structural features maximise the rate of ATP production.

Question 6: Compare the structure and function of DNA and mRNA.

Structure: DNA is double-stranded with a deoxyribose sugar, thymine as a base, and is very long. MRNA is single-stranded with a ribose sugar, uracil replacing thymine, and is much shorter.

Function: DNA stores genetic information long-term in the nucleus. MRNA carries a copy of the Genetic information from the nucleus to the ribosomes for protein synthesis (transient role).

Question 7: Describe the stages of mitosis and explain their significance.

Prophase: Chromosomes condense (become visible), the nuclear envelope breaks down, spindle Fibres form, and centrioles move to opposite poles. This prepares the cell for division by Organising the genetic material.

Metaphase: Chromosomes align at the metaphase plate (cell equator) with spindle fibres attached To centromeres. This ensures equal distribution of chromosomes to daughter cells.

Anaphase: Sister chromatids separate at the centromeres and are pulled to opposite poles by Shortening spindle fibres. This ensures each daughter cell receives identical genetic material.

Telophase: Chromosomes decondense, nuclear envelopes reform, nucleoli reappear, and spindle Fibres break down. This reverses the changes of prophase and re-establishes the interphase state.

Cytokinesis: The cytoplasm divides, producing two separate daughter cells.

Question 8: A cell has 20 chromosomes in G1. How many chromosomes and DNA molecules are present At the end of S phase, during metaphase of mitosis, and after cytokinesis?

End of S phase: 20 chromosomes, 40 DNA molecules (each chromosome has been replicated into two Sister chromatids).

Metaphase of mitosis: 20 chromosomes (each with 2 chromatids), 40 DNA molecules.

After cytokinesis: 20 chromosomes, 20 DNA molecules (each daughter cell receives 20 single-chromatid Chromosomes).


Cells communicate through chemical messengers called ligands that bind to specific receptors.

Types of cell signalling:

TypeDescriptionDistanceExample
AutocrineCell signals to itselfSame cellGrowth factors
ParacrineCell signals to nearby cellsShort distanceNeurotransmitters
EndocrineHormone travels in bloodLong distanceInsulin, adrenaline
SynapticSignal across a synapseVery shortAcetylcholine
Contact-dependentDirect cell-to-cell contactAdjacent cellsImmune recognition

Cell-surface receptors: For large or hydrophilic signalling molecules that cannot cross the cell Membrane (e.g., insulin, adrenaline).

Intracellular receptors: For small or hydrophobic signalling molecules that can diffuse through The membrane (e.g., steroid hormones like testosterone, oestrogen).

When a ligand binds to a cell-surface receptor, it triggers a cascade of intracellular events:

  1. Signal reception: The ligand binds to the receptor
  2. Signal transduction: The signal is relayed inside the cell, often involving:
  • G-proteins: Activate or inhibit enzymes
  • Second messengers: Small molecules that amplify the signal (e.g., cAMP, Ca2+\mathrm{Ca}^{2+})
  • Enzyme cascades: Kinase cascades that phosphorylate target proteins
  1. Cellular response: The cell changes its activity (e.g., gene expression, metabolism, secretion)

First line of defence:

  • Skin: physical barrier
  • Mucous membranes: trap pathogens
  • Stomach acid: destroys pathogens
  • Tears, saliva: contain lysozyme (an enzyme that breaks down bacterial cell walls)

Second line of defence:

  • Phagocytes (neutrophils, macrophages): engulf and digest pathogens by phagocytosis
  • Inflammation: increased blood flow, swelling, heat, pain; brings phagocytes to the infection site
  • Fever: raises body temperature, inhibiting pathogen growth and enhancing immune function
  • Interferons: proteins produced by virus-infected cells that inhibit viral replication in neighbouring cells

Cell-mediated response (T cells):

  • T helper cells: release cytokines that stimulate B cells and cytotoxic T cells
  • Cytotoxic T cells (killer T cells): destroy virus-infected cells and cancer cells by inducing apoptosis
  • Memory T cells: provide long-term immunity

Humoral response (B cells):

  • B cells produce antibodies (immunoglobulins) specific to the antigen
  • Plasma cells: short-lived cells that secrete large quantities of antibodies
  • Memory B cells: provide long-term immunity; respond faster and more strongly upon re-exposure

An antibody (immunoglobulin) is a Y-shaped protein with:

  • Two identical heavy chains
  • Two identical light chains
  • Variable regions: specific to the antigen (at the tips of the Y)
  • Constant regions: the same for all antibodies of the same class
TypeLocationFunction
IgGBlood, tissue fluidMost abundant; crosses placenta
IgASaliva, tears, breast milk, mucusProtects mucous membranes
IgMBloodFirst antibody produced in primary response
IgEBound to mast cellsAllergic reactions; parasitic infections
IgDB cell surfaceB cell activation
FeaturePrimary ResponseSecondary Response
SpeedSlow (5-10 days)Fast (1-3 days)
Antibody levelLowerHigher
Antibody classMainly IgMMainly IgG
DurationShortLong
Memory cellsProducedAlready present

Explain how vaccination provides immunity against a disease.

A vaccine contains a weakened or dead form of the pathogen (or parts of it like antigens). When Introduced into the body, it triggers a primary immune response without causing the disease. B cells Produce antibodies and memory B cells are formed. If the person is later exposed to the actual Pathogen, the memory B cells quickly produce large quantities of antibodies in a secondary response, Destroying the pathogen before symptoms develop. This provides active artificial immunity.


Definition: Undifferentiated cells that can divide and differentiate into specialised cell Types.

TypeSourcePotential
TotipotentEarly embryo (up to 8-cell stage)Can become any cell type + placenta
PluripotentBlastocyst (inner cell mass)Can become any cell type
MultipotentAdult tissues (e.g., bone marrow)Can become limited range of cell types
UnipotentSpecific tissuesCan become only one cell type

Therapeutic uses:

  • Treatment of leukaemia (bone marrow transplants)
  • Potential for regenerative medicine (repairing damaged tissues)
  • Research into disease mechanisms

Ethical considerations:

  • Embryonic stem cells involve destruction of embryos
  • Adult stem cells have more limited potential
  • Induced pluripotent stem cells (iPSCs) offer an alternative

The water potential of a solution is:

Ψ=Ψs+Ψp\Psi = \Psi_s + \Psi_p

Where:

  • Ψs\Psi_s = solute potential (always negative or zero)
  • Ψp\Psi_p = pressure potential (positive in turgid plant cells, zero in animal cells)

For a solution with no pressure applied:

Ψ=Ψs=iCRT\Psi = \Psi_s = -iCRT

Where:

  • ii = ionisation constant (number of particles per molecule)
  • CC = molar concentration
  • RR = gas constant (8.314Jmol1K18.314 \mathrm{ J mol}^{-1} \mathrm{ K}^{-1})
  • TT = temperature in Kelvin

Calculate the water potential of a 0.3mol/dm30.3 \mathrm{ mol/dm}^3 sucrose solution at 25C25^\circ\mathrm{C}. (i=1i = 1 for sucrose)

Ψs=iCRT=(1)(0.3)(8.314)(298)=743.3kPa\Psi_s = -iCRT = -(1)(0.3)(8.314)(298) = -743.3 \mathrm{ kPa}

Since there is no pressure: Ψ=743.3kPa\Psi = -743.3 \mathrm{ kPa}

Calculate the water potential of a 0.2mol/dm30.2 \mathrm{ mol/dm}^3 NaCl\mathrm{NaCl} solution at 20C20^\circ\mathrm{C}. (i=2i = 2 for NaCl\mathrm{NaCl})

Ψs=(2)(0.2)(8.314)(293)=974.4kPa\Psi_s = -(2)(0.2)(8.314)(293) = -974.4 \mathrm{ kPa}

Ψ=974.0kPa\Psi = -974.0 \mathrm{ kPa}


Problem 1: Describe three structural differences between prokaryotic and eukaryotic cells, and explain how each difference relates to the cell’s function.

If you get this wrong, revise: Cell Structure — Prokaryotic vs Eukaryotic Cells

Solution
  1. Nucleus: Eukaryotic cells have a true nucleus with a nuclear envelope, while prokaryotic cells have a nucleoid region (no membrane). The nuclear envelope in eukaryotes separates transcription from translation, allowing mRNA processing (splicing, capping, poly-A tail), enabling more complex gene regulation.

  2. Mitochondria: Eukaryotic cells have mitochondria for aerobic respiration (high ATP yield via oxidative phosphorylation), while prokaryotic cells carry out respiration on their cell membrane (less efficient, limited surface area). Mitochondria provide a large internal surface area (cristae) for the electron transport chain.

  3. Ribosomes: Eukaryotic cells have 80S ribosomes (larger), while prokaryotic cells have 70S ribosomes (smaller). This difference is the basis of antibiotic specificity — drugs like tetracycline target 70S ribosomes in bacteria without affecting eukaryotic 80S ribosomes.

Problem 2: A red blood cell is placed in a 0.5 mol/dm3^3 sucrose solution. Predict what will happen to the cell and explain your reasoning. (The water potential of the red blood cell cytoplasm is approximately -700 kPa.)

If you get this wrong, revise: Membrane Transport — Osmosis; Water Potential and Osmosis Calculations

Solution

The water potential of the 0.5 mol/dm3^3 sucrose solution is:

Ψs=iCRT=(1)(0.5)(8.314)(298)=1238.8kPa\Psi_s = -iCRT = -(1)(0.5)(8.314)(298) = -1238.8 \mathrm{ kPa}

Since the solution has a more negative water potential (-1238.8 kPa) than the cell cytoplasm (-700 kPa), water will move out of the cell by osmosis (from higher to lower water potential). The red blood cell will lose water, shrink, and become crenated (wrinkled appearance).

Problem 3: Explain the effect of increasing temperature on enzyme activity from 0 degrees C to 60 degrees C. Refer to kinetic energy, collision frequency, and enzyme structure in your answer.

If you get this wrong, revise: Enzymes — Factors Affecting Enzyme Activity

Solution

From 0 degrees C to the optimum (~37-40 degrees C): increasing temperature increases the kinetic energy of both enzyme and substrate molecules, leading to more frequent effective collisions between them. The rate of reaction increases as more enzyme-substrate complexes form per unit time.

Beyond the optimum temperature, the increased kinetic energy breaks the hydrogen bonds, ionic bonds, and other weak interactions maintaining the tertiary structure of the enzyme. The active site changes shape (denaturation), and the substrate can no longer bind. Since denaturation is irreversible, the reaction rate decreases sharply and the enzyme is permanently inactivated.

Problem 4: A cell in the G1 phase of the cell cycle has 46 chromosomes. How many chromosomes and DNA molecules are present at the end of S phase, during metaphase of mitosis, and after cytokinesis?

If you get this wrong, revise: Cell Division — Mitosis; The Cell Cycle

Solution

End of S phase: 46 chromosomes, 92 DNA molecules (each chromosome has been replicated into two sister chromatids, but chromosome count remains 46 because sister chromatids are still joined at the centromere).

Metaphase of mitosis: 46 chromosomes (each with 2 chromatids), 92 DNA molecules. The chromosomes are aligned at the metaphase plate with spindle fibres attached to centromeres.

After cytokinesis: 46 chromosomes, 46 DNA molecules. Each daughter cell receives 46 single-chromatid chromosomes (the chromatids separated during anaphase). The chromosome number is the same as the original cell, maintaining genetic continuity.

Problem 5: Compare the structure and function of DNA and mRNA.

If you get this wrong, revise: DNA Replication; Protein Synthesis

Solution

Structure: DNA is double-stranded (double helix) with a deoxyribose sugar and thymine as a base. MRNA is single-stranded with a ribose sugar and uracil replacing thymine. DNA is very long (entire genome), while mRNA is a shorter copy of a single gene.

Function: DNA stores genetic information long-term in the nucleus. MRNA carries a transient copy of the genetic information from the nucleus to the ribosomes in the cytoplasm for protein synthesis (translation). DNA is self-replicating; mRNA is not.

Stability: DNA is chemically stable (deoxyribose is less reactive than ribose; double-stranded structure provides protection). MRNA is short-lived (ribose is more reactive; single-stranded structure is vulnerable to nucleases), allowing rapid changes in gene expression.

Problem 6: Describe the stages of mitosis and explain their significance in ensuring genetic continuity.

If you get this wrong, revise: Cell Division — Stages of Mitosis

Solution

Prophase: Chromatin condenses into visible chromosomes (each consisting of two sister chromatids joined at the centromere). The nuclear envelope breaks down, spindle fibres form, and centrioles move to opposite poles. Significance: organises and packages the genetic material for even distribution.

Metaphase: Chromosomes align at the metaphase plate (cell equator) with spindle fibres attached to centromeres. Significance: ensures each daughter cell receives one copy of each chromosome.

Anaphase: Sister chromatids separate at the centromeres and are pulled to opposite poles by shortening spindle fibres. Significance: the critical step that distributes identical genetic material to each daughter cell.

Telophase: Chromosomes decondense, nuclear envelopes reform, nucleoli reappear, and spindle fibres break down. Significance: re-establishes the interphase state in each daughter cell.

Cytokinesis: The cytoplasm divides, producing two separate daughter cells, each genetically identical to the parent cell.

Problem 7: Explain how vaccination provides immunity against a disease, referring to both the primary and secondary immune responses.

If you get this wrong, revise: Immune System — Specific (Adaptive) Immunity; Primary vs Secondary Immune Response

Solution

A vaccine contains a weakened or dead form of the pathogen (or specific antigens). When introduced into the body, it triggers a primary immune response: B cells are activated by the antigen, divide by mitosis, and differentiate into plasma cells (which secrete antibodies) and memory B cells. T helper cells are also activated and support the B cell response.

If the person is later exposed to the actual pathogen, memory B cells recognise the antigen and rapidly divide, producing large quantities of antibodies in a secondary response. The secondary response is faster (1-3 days vs 5-10 days), produces more antibodies (mainly IgG), and lasts longer. The pathogen is destroyed before it can cause symptoms, providing active artificial immunity.

Problem 8: Describe the role of the Golgi apparatus in protein processing and transport.

If you get this wrong, revise: Cell Organelles — Golgi Apparatus

Solution

The Golgi apparatus receives proteins from the rough ER in transport vesicles. It modifies these proteins by adding carbohydrate groups (glycosylation) to form glycoproteins. It sorts proteins based on their destination and packages them into secretory vesicles. Different vesicles are targeted to different locations: some fuse with the cell membrane for export (exocytosis), some become lysosomes (in animal cells), and others are transported to other organelles. The Golgi acts as a processing and distribution centre, ensuring proteins reach their correct destinations.

Problem 9: Calculate the water potential of a 0.15 mol/dm3^3 NaCl solution at 25 degrees C. (i=2i = 2 for NaCl.) Determine whether a plant cell with Ψ=800kPa\Psi = -800 \mathrm{ kPa} and Ψp=200kPa\Psi_p = 200 \mathrm{ kPa} would gain or lose water in this solution.

If you get this wrong, revise: Water Potential and Osmosis Calculations

Solution

Ψs=iCRT=(2)(0.15)(8.314)(298)=742.5kPa\Psi_s = -iCRT = -(2)(0.15)(8.314)(298) = -742.5 \mathrm{ kPa}

Since there is no pressure applied to the external solution: Ψsolution=742.5kPa\Psi_{\mathrm{solution}} = -742.5 \mathrm{ kPa}

For the plant cell:

Ψcell=Ψs+Ψp=800+200=600kPa\Psi_{\mathrm{cell}} = \Psi_s + \Psi_p = -800 + 200 = -600 \mathrm{ kPa}

Since Ψsolution\Psi_{\mathrm{solution}} (-742.5 kPa) is more negative than Ψcell\Psi_{\mathrm{cell}} (-600 kPa), water moves from the cell (higher water potential) into the solution (lower water potential). The plant cell would lose water.

Problem 10: Explain how the structure of the mitochondrion is adapted to its function in aerobic respiration.

If you get this wrong, revise: Cell Organelles — Mitochondria; Cellular Respiration

Solution

The mitochondrion has a double membrane. The inner membrane is folded into cristae, providing a large surface area for the electron transport chain and ATP synthase. The intermembrane space between the two membranes allows protons to accumulate, creating the electrochemical gradient needed for chemiosmosis. The matrix contains enzymes for the Krebs cycle, mitochondrial DNA (allowing the mitochondrion to produce some of its own proteins), and ribosomes. The small volume of the matrix ensures high concentrations of substrates and enzymes, maximising the rate of the Krebs cycle. These features collectively maximise the rate of ATP production through oxidative phosphorylation.


Glycolysis occurs in the cytoplasm and does not require oxygen. It is the first stage of both aerobic and anaerobic respiration.

Stages of glycolysis:

StageKey StepsEnergy Change
PhosphorylationGlucose (6C) is phosphorylated twice using 2 ATP, forming hexose biphosphate (6C). This “traps” glucose inside the cell and raises its energy levelConsumes 2 ATP (investment phase)
SplittingHexose biphosphate is split into two molecules of triose phosphate (3C each)No ATP change
OxidationEach triose phosphate is oxidised (dehydrogenated) by NAD+^+Producing 2 NADH (one per triose phosphate)Produces 2 NADH
ATP productionEach triose phosphate is converted to pyruvate (3C), producing 2 ATP per triose phosphate (substrate-level phosphorylation)Produces 4 ATP (2 per triose phosphate)
Net yield2 ATP + 2 NADH per glucose

Glucose (6C)+2NAD++2ADP+2Pi2 Pyruvate (3C)+2NADH+2H++2ATP+2H2O\text{Glucose (6C)} + 2\mathrm{NAD}^+ + 2\mathrm{ADP} + 2\mathrm{P}_i \to 2\text{ Pyruvate (3C)} + 2\mathrm{NADH} + 2\mathrm{H}^+ + 2\mathrm{ATP} + 2\mathrm{H}_2\mathrm{O}

Link reaction (pyruvate oxidation):

  • Occurs in the mitochondrial matrix
  • Each pyruvate (3C) is decarboxylated (loses CO2\mathrm{CO}_2) and dehydrogenated (NAD+^+ reduced to NADH)
  • The remaining 2-carbon acetyl group combines with coenzyme A to form acetyl CoA
  • Per glucose: 2 pyruvate \to 2 acetyl CoA + 2 CO2\mathrm{CO}_2 + 2 NADH

Krebs cycle (citric acid cycle):

  • Occurs in the mitochondrial matrix
  • Each acetyl CoA (2C) combines with oxaloacetate (4C) to form citrate (6C)
  • Through a series of reactions, citrate is converted back to oxaloacetate, releasing:
  • 2 CO2\mathrm{CO}_2 (decarboxylation)
  • 3 NADH (dehydrogenation)
  • 1 FADH2_2 (dehydrogenation)
  • 1 ATP (substrate-level phosphorylation via GTP)
  • Per glucose: 2 turns of the cycle produce 4 CO2\mathrm{CO}_26 NADH, 2 FADH_2\_22 ATP
  • Occurs on the inner mitochondrial membrane (cristae)
  • NADH and FADH2_2 donate electrons to the electron transport chain
  • Electrons pass through a series of carriers (complexes I-IV), releasing energy used to pump H+\mathrm{H}^+ into the intermembrane space
  • H+\mathrm{H}^+ flows back through ATP synthase (chemiosmosis), producing ATP
  • Oxygen is the final electron acceptor, combining with H+\mathrm{H}^+ and electrons to form water
StageATP (or equivalent)
Glycolysis (net)2 ATP + 2 NADH
Link reaction2 NADH
Krebs cycle2 ATP + 6 NADH + 2 FADH2_2
Oxidative phosphorylationEach NADH produces approximately 2.5 ATP; each FADH2_2 produces approximately 1.5 ATP
Total (approximate)30-32 ATP