Cell Biology

1. Cell Theory

Postulates of Cell Theory

Cell theory, formulated by Schleiden and Schwann (1838--1839) and later refined by Virchow, rests on three principles:

1. All living organisms are composed of one or more cells.
2. The cell is the smallest unit of life.
3. All cells arise from pre-existing cells (biogenesis), refuting spontaneous generation.

Definition. Unicellular organisms carry out all life functions within a single cell. Multicellular organisms have differentiated cells that specialise in specific functions.

Evidence for Cell Theory

• Microscopy: light and electron microscopy confirm that all examined living tissue is cellular.
• Biogenesis experiments: Pasteur's swan-neck flask experiment (1859) demonstrated that microorganisms arise only from existing microorganisms, not from non-living matter.
• Cell culture: individual cells can be grown in vitro and give rise to colonies, confirming that single cells are viable living units.

Exceptions and Caveats

• Viruses are acellular and require a host cell to replicate; they are not considered living organisms under cell theory.
• The first cell must have arisen from non-cellular precursors (abiogenesis), which cell theory does not address.

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2. Cell Types

Prokaryotic Cells

Prokaryotes (bacteria and archaea) lack a membrane-bound nucleus. Their DNA is found as a single, circular chromosome in the nucleoid region.

Feature	Description
Cell wall	Peptidoglycan (bacteria) or pseudopeptidoglycan (archaea)
Plasma membrane	Phospholipid bilayer with embedded proteins
Nucleoid	Region containing the circular chromosome; not enclosed by a membrane
Ribosomes	$70\mathrm{S}$ ($50\mathrm{S}$ + $30\mathrm{S}$ subunits)
Plasmids	Small circular extra-chromosomal DNA molecules
Flagella	Rotatory propulsion structures (not present in all species)
Pili	Hair-like appendages for adhesion and conjugation
Capsule / slime layer	Polysaccharide layer for protection and adhesion

Eukaryotic Cells

Eukaryotes (animals, plants, fungi, protists) possess a membrane-bound nucleus and organelles.

Feature	Animal Cell	Plant Cell
Cell wall	Absent	Cellulose and other polysaccharides
Chloroplasts	Absent	Present
Large central vacuole	Small, temporary	Large, permanent, maintains turgor
Centrioles	Present (for spindle formation)	Absent
Lysosomes	Present	Rare or absent (vacuoles serve role)
Plastids	Absent	Present

Comparison

Feature	Prokaryote	Eukaryote
Nucleus	Absent	Present
DNA form	Circular	Linear chromosomes
Membrane-bound organelles	None	Many (mitochondria, ER, Golgi, etc.)
Ribosome size	$70\mathrm{S}$	$80\mathrm{S}$ ($60\mathrm{S}$ + $40\mathrm{S}$)
Cell diameter	$1$--$5\;\mathrm{\mu m}$	$10$--$100\;\mathrm{\mu m}$
Reproduction	Binary fission	Mitosis / meiosis

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3. Cell Membrane Structure

The Fluid Mosaic Model

The cell membrane is described by the fluid mosaic model (Singer and Nicolson, 1972). The membrane is a dynamic, two-dimensional fluid of phospholipids in which proteins are embedded ("mosaic").

Structure:

• Phospholipid bilayer: hydrophilic phosphate heads face the aqueous environment on both sides; hydrophobic fatty acid tails form the interior. This creates a selectively permeable barrier.
• Cholesterol: wedged between phospholipids in animal cells, modulating membrane fluidity. At low temperatures it prevents packing; at high temperatures it restrains movement.
• Integral (transmembrane) proteins: span the bilayer and function as channels, carriers, or receptors.
• Peripheral proteins: attached to one surface of the membrane, involved in signalling or cytoskeletal anchoring.
• Glycoproteins / glycolipids: carbohydrate chains on the extracellular surface for cell recognition and signalling.

Membrane Transport

Simple Diffusion

Passive movement of molecules from a region of higher concentration to lower concentration, down the concentration gradient. Only small, non-polar molecules ($\mathrm{O}_2$, $\mathrm{CO}_2$, steroid hormones) diffuse directly through the bilayer.

Fick's Law:

$$\mathrm{Rate of diffusion} \propto \frac{(\mathrm{surface area}) \times (\mathrm{concentration difference})}{\mathrm{thickness of membrane}}$$

Facilitated Diffusion

Polar or charged molecules (glucose, ions, amino acids) require transmembrane proteins. No ATP is consumed; transport is passive and specific.

• Channel proteins: form hydrophilic pores (e.g., aquaporins for water).
• Carrier proteins: undergo conformational change to shuttle substances across.

Osmosis

Definition. Osmosis is the net movement of water molecules across a selectively permeable membrane from a region of lower solute concentration (higher water potential) to a region of higher solute concentration (lower water potential).

Water potential ($\psi$) is measured in pressure units ($\mathrm{kPa}$). Pure water at standard conditions has $\psi = 0\;\mathrm{kPa}$. Adding solutes decreases $\psi$ (makes it negative).

Solution type	Effect on animal cell	Effect on plant cell
Hypotonic	Swells, may lyse (crenation opposite)	Turgid (normal, healthy)
Isotonic	No net change	Flaccid
Hypertonic	Crenates (shrinks)	Plasmolysed

Active Transport

Movement against the concentration gradient, requiring ATP directly or indirectly.

• Primary active transport: ATP hydrolysis drives the transport protein directly (e.g., the $\mathrm{Na}^+/\mathrm{K}^+$ pump moves $3\;\mathrm{Na}^+$ out and $2\;\mathrm{K}^+$ in per ATP).
• Secondary active transport (co-transport): uses the electrochemical gradient established by primary active transport. E.g., $\mathrm{Na}^+$-glucose co-transport in intestinal epithelium.

Vesicular Transport

• Endocytosis: bulk uptake of material by invagination of the membrane.
  • Phagocytosis: solid particles (e.g., white blood cells engulfing bacteria).
  • Pinocytosis: liquid droplets.
• Exocytosis: vesicles fuse with the plasma membrane to release contents (e.g., secretion of hormones, neurotransmitters).

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4. Organelles

Nucleus

The nucleus houses the genetic material (DNA) and controls cellular activities through gene expression.

• Nuclear envelope: double membrane with nuclear pores (diameter $\approx 9\;\mathrm{nm}$) allowing selective transport of mRNA, proteins, and ribosomal subunits.
• Nucleolus: site of rRNA transcription and ribosome subunit assembly.
• Chromatin: DNA wrapped around histone proteins; condenses into chromosomes during cell division.

Mitochondria

Site of aerobic respiration (oxidative phosphorylation), producing ATP.

• Double membrane: outer membrane is permeable; inner membrane is folded into cristae to increase surface area.
• Matrix: contains its own circular DNA ($70\mathrm{S}$ ribosomes), enzymes for the Krebs cycle, and the link reaction.
• The endosymbiotic theory posits that mitochondria evolved from free-living aerobic prokaryotes engulfed by ancestral eukaryotes.

Endoplasmic Reticulum

• Rough ER (RER): studded with ribosomes; synthesises proteins destined for secretion, membranes, or organelles. Proteins enter the ER lumen and undergo folding and post-translational modification.
• Smooth ER (SER): lacks ribosomes; involved in lipid synthesis, detoxification (liver cells), and calcium storage (muscle cells).

Golgi Apparatus

A stack of flattened, membrane-bound cisternae that modifies, sorts, and packages proteins and lipids received from the ER.

• Cis face (receiving side): accepts vesicles from the ER.
• Trans face (shipping side): releases vesicles to the plasma membrane, lysosomes, or for secretion.
• Glycosylation (addition of sugar groups) occurs in the Golgi.

Lysosomes

Membrane-bound vesicles containing hydrolytic enzymes (proteases, lipases, nucleases) active at $\mathrm{pH}\approx 5$.

• Digest worn-out organelles (autophagy).
• Break down materials taken in by phagocytosis.
• Absent from most plant cells (central vacuole performs analogous function).

Ribosomes

Sites of protein synthesis (translation). Composed of rRNA and proteins.

• Free ribosomes: in the cytoplasm; synthesise proteins used within the cell.
• Bound ribosomes: attached to the RER; synthesise proteins for secretion or membranes.

Cell Wall (Plants)

Made primarily of cellulose microfibrils embedded in a matrix of hemicellulose and pectin. Provides structural support and prevents osmotic lysis.

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5. Cell Division

Mitosis

Mitosis produces two genetically identical diploid ($2n$) daughter cells, maintaining the chromosome number. It is the basis of growth, repair, and asexual reproduction.

Definition. A diploid cell contains two complete sets of chromosomes ($2n$). A haploid cell contains one set ($n$).

Definition. Homologous chromosomes are a pair of chromosomes (one maternal, one paternal) of the same length, centromere position, and gene loci.

Definition. Sister chromatids are two identical copies of a single chromosome, joined at the centromere, produced by DNA replication in S phase.

Phases of Mitosis

Phase	Key Events
Prophase	Chromatin condenses into visible chromosomes; spindle microtubules begin to form from the centrosomes; nucleolus disappears.
Metaphase	Chromosomes align at the metaphase plate (cell equator); spindle fibres attach to kinetochores at centromeres.
Anaphase	Sister chromatids separate at the centromere and are pulled to opposite poles by shortening spindle fibres. Chromatids are now individual chromosomes.
Telophase	Chromosomes decondense; nuclear envelope reforms around each set of chromosomes; nucleolus reappears.

Cytokinesis (division of the cytoplasm) follows mitosis:

• Animals: a cleavage furrow forms (actin ring constricts).
• Plants: a cell plate forms from vesicles at the equator, becoming the new cell wall.

The Cell Cycle

The cell cycle has four phases:

1. $\mathrm{G}_1$ (Gap 1): cell growth, organelle duplication.
2. S (Synthesis): DNA replication; each chromosome becomes two sister chromatids.
3. $\mathrm{G}_2$ (Gap 2): preparation for mitosis; protein synthesis.
4. M (Mitotic): mitosis and cytokinesis.

$\mathrm{G}_0$ is a resting phase outside the cycle; some cells (neurones) enter $\mathrm{G}_0$ permanently.

Meiosis

Meiosis produces four genetically distinct haploid ($n$) daughter cells from one diploid ($2n$) parent cell. It is essential for sexual reproduction, producing gametes (sperm and egg cells).

Meiosis I (Reductional Division)

Phase	Key Events
Prophase I	Homologous chromosomes pair up (synapsis) forming bivalents; crossing over occurs at chiasmata, exchanging genetic material between non-sister chromatids.
Metaphase I	Bivalents align at the metaphase plate; homologous pairs (not individual chromosomes) are positioned.
Anaphase I	Homologous chromosomes separate and move to opposite poles. Sister chromatids remain attached.
Telophase I	Two haploid cells form (each chromosome still consists of two chromatids).

Meiosis II (Equational Division)

Similar to mitosis but with haploid cells:

Phase	Key Events
Prophase II	Chromosomes condense again; spindle forms.
Metaphase II	Individual chromosomes align at the metaphase plate.
Anaphase II	Sister chromatids separate and move to opposite poles.
Telophase II	Four haploid daughter cells form, each genetically distinct.

Sources of Genetic Variation

1. Crossing over (Prophase I): recombination between non-sister chromatids produces new allele combinations on chromosomes.
2. Independent assortment (Metaphase I): the orientation of each bivalent is random; $2^n$ possible chromosome combinations (where $n$ is the haploid number).
3. Random fertilisation: any sperm can fuse with any egg, further increasing variability.

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6. Specialised Cells

Multicellular organisms have differentiated (specialised) cell types:

Cell type	Adaptation	Function
Red blood cell	Biconcave disc, no nucleus, high haemoglobin	$\mathrm{O}_2$ transport
Sperm cell	Flagellum, many mitochondria, acrosome	Fertilisation
Neuron	Long axon, dendrites, myelin sheath	Signal transmission
Muscle cell	Many mitochondria, contractile filaments	Contraction
Root hair cell	Elongated projection, large surface area	Mineral and water uptake
Palisade mesophyll	Many chloroplasts, elongated shape	Photosynthesis

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Common Pitfalls

• Confusing chromosomes and chromatids: a replicated chromosome consists of two sister chromatids joined at a centromere. After anaphase, each chromatid is an individual chromosome.
• Confusing mitosis and meiosis: mitosis produces $2$ identical diploid cells; meiosis produces $4$ distinct haploid cells.
• Equating osmosis with diffusion: osmosis refers specifically to water movement across a selectively permeable membrane.
• Stating that prokaryotes "have no DNA": they possess a circular chromosome and plasmids, but lack a membrane-bound nucleus.
• Describing the cell membrane as a "solid" structure: it is fluid, with phospholipids and proteins able to move laterally.

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Practice Problems

Question 1: Prokaryote vs Eukaryote Ribosomes

Explain why the antibiotic streptomycin can inhibit protein synthesis in bacteria without affecting human cells. Reference ribosome structure in your answer.

Answer

Streptomycin binds specifically to the $30\mathrm{S}$ subunit of prokaryotic $70\mathrm{S}$ ribosomes, disrupting translation. Eukaryotic ribosomes are $80\mathrm{S}$ ($60\mathrm{S}$ + $40\mathrm{S}$ subunits), which have a different structure and do not bind streptomycin. This structural difference allows selective toxicity against bacterial cells.

Question 2: Surface Area to Volume Ratio

A spherical cell has a radius of $5\;\mathrm{\mu m}$. Calculate its surface area, volume, and surface-area-to-volume ratio. Explain why cells are generally small.

Answer

For a sphere: $A = 4\pi r^2$, $V = \frac{4}{3}\pi r^3$.

$A = 4\pi(5)^2 = 100\pi \approx 314\;\mathrm{\mu m}^2$

$V = \frac{4}{3}\pi(5)^3 = \frac{500\pi}{3} \approx 524\;\mathrm{\mu m}^3$

$\frac{A}{V} = \frac{3}{r} = \frac{3}{5} = 0.6\;\mathrm{\mu m}^{-1}$

Cells are small because the surface-area-to-volume ratio decreases as cell size increases. A lower ratio means insufficient surface area for nutrient uptake and waste removal relative to the metabolic demands of the cell volume.

Question 3: Osmosis Calculation

An animal cell with a water potential of $-700\;\mathrm{kPa}$ is placed in a solution with a water potential of $-400\;\mathrm{kPa}$. Predict the net movement of water and describe the effect on the cell.

Answer

Water moves from higher water potential ($-400\;\mathrm{kPa}$) to lower water potential ($-700\;\mathrm{kPa}$). The net movement of water is into the cell by osmosis. The cell will swell as water enters, and if the volume increase is sufficient, the cell may undergo lysis (burst), since animal cells lack a rigid cell wall.

Question 4: Meiosis and Genetic Variation

A diploid organism has $2n = 8$ chromosomes. How many genetically distinct gametes can be produced from independent assortment alone? If a single pair of homologous chromosomes undergoes a single crossover, how many chromatids are affected?

Answer

With $2n = 8$, the haploid number is $n = 4$. Independent assortment alone yields $2^n = 2^4 = 16$ possible chromosome combinations.

A single crossover between non-sister chromatids of one homologous pair affects exactly two (non-sister) chromatids: one from each homologous chromosome.

Question 5: Identifying Organelles from Micrographs

An electron micrograph shows a double-membrane organelle with inner folds and a dense matrix containing DNA. Identify the organelle and explain two features visible in the micrograph that support its function.

Answer

The organelle is a mitochondrion. The inner folds (cristae) increase the surface area for the electron transport chain and ATP synthase, maximising ATP production. The presence of its own DNA supports the endosymbiotic theory and allows the mitochondrion to produce some of its own proteins independently of nuclear genes.

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Worked Examples

Worked Example: Surface Area-to-Volume Ratio and Cell Size Limit

A cuboidal epithelial cell has a side length of $10\;\mathrm{\mu m}$. It doubles in each linear dimension to become $20\;\mathrm{\mu m}$ per side. Calculate the surface-area-to-volume ratio before and after the enlargement, and determine the factor by which the ratio decreases.

Solution

For a cube: $A = 6s^2$ and $V = s^3$.

Before enlargement ($s = 10\;\mathrm{\mu m}$): $A = 6 \times 10^2 = 600\;\mathrm{\mu m}^2$ $V = 10^3 = 1000\;\mathrm{\mu m}^3$ $\frac{A}{V} = \frac{600}{1000} = 0.60\;\mathrm{\mu m}^{-1}$

After enlargement ($s = 20\;\mathrm{\mu m}$): $A = 6 \times 20^2 = 2400\;\mathrm{\mu m}^2$ $V = 20^3 = 8000\;\mathrm{\mu m}^3$ $\frac{A}{V} = \frac{2400}{8000} = 0.30\;\mathrm{\mu m}^{-1}$

The surface-area-to-volume ratio is halved. This demonstrates why larger cells face a fundamental constraint: the metabolic demand (proportional to volume) grows faster than the capacity for nutrient and waste exchange (proportional to surface area). Multicellular organisms compensate by folding membranes (microvilli, cristae) to increase effective surface area without increasing cell volume.

Worked Example: Magnification Calculation from a Micrograph

An electron micrograph of a chloroplast shows a granum that measures $4.2\;\mathrm{cm}$ across the image. The scale bar indicates that $1\;\mathrm{cm}$ on the image represents $0.5\;\mathrm{\mu m}$ in reality. Calculate the actual width of the granum and the magnification of the micrograph.

Solution

Actual width: $4.2\;\mathrm{cm} \times 0.5\;\mathrm{\mu m/cm} = 2.1\;\mathrm{\mu m}$

Magnification: $\mathrm{Magnification} = \frac{\mathrm{image size}}{\mathrm{actual size}} = \frac{4.2\;\mathrm{cm}}{2.1\;\mathrm{\mu m}}$

Convert to consistent units: $4.2\;\mathrm{cm} = 42000\;\mathrm{\mu m}$

$\mathrm{Magnification} = \frac{42000}{2.1} = 20000\times$

The granum is $2.1\;\mathrm{\mu m}$ wide and the micrograph is magnified $20000$ times. A typical granum stack contains $5$--$30$ thylakoid discs, each approximately $0.3\;\mathrm{\mu m}$ in diameter, consistent with this measurement.

Worked Example: Interpreting the Na+/K+ Pump Stoichiometry

The $\mathrm{Na}^+/\mathrm{K}^+$ ATPase pumps $3\;\mathrm{Na}^+$ ions out and $2\;\mathrm{K}^+$ ions in per ATP hydrolysed. A neurone at rest has an intracellular $\mathrm{Na}^+$ concentration of $15\;\mathrm{mmol/L}$ and an extracellular concentration of $145\;\mathrm{mmol/L}$. Calculate the ratio of the concentration gradients and explain why the pump is described as electrogenic.

Solution

$\mathrm{Na}^+$ concentration gradient: $\frac{[\mathrm{Na}^+]_{\mathrm{out}}}{[\mathrm{Na}^+]_{\mathrm{in}}} = \frac{145}{15} \approx 9.7$

$\mathrm{K}^+$ concentration gradient (typical values: $150\;\mathrm{mmol/L}$ in, $5\;\mathrm{mmol/L}$ out): $\frac{[\mathrm{K}^+]_{\mathrm{in}}}{[\mathrm{K}^+]_{\mathrm{out}}} = \frac{150}{5} = 30$

The pump is electrogenic because it moves $3$ positive charges out but only $2$ positive charges in per cycle, resulting in a net export of one positive charge per ATP hydrolysed. This net outward current contributes directly to the negative resting membrane potential (approximately $-70\;\mathrm{mV}$), alongside the $\mathrm{K}^+$ diffusion potential established by leak channels. If the pump moved equal numbers of cations in each direction, it would not directly contribute to the membrane potential.

Worked Example: Counting Cells from a Haemocytometer

A student loads a haemocytometer with a yeast cell suspension. The central large square ($1\;\mathrm{mm} \times 1\;\mathrm{mm}$, depth $0.1\;\mathrm{mm}$) is divided into $25$ smaller squares. The student counts an average of $42$ yeast cells per small square across $5$ squares. Calculate the cell concentration in cells per $\mathrm{mL}$.

Solution

Volume of the central large square: $1\;\mathrm{mm} \times 1\;\mathrm{mm} \times 0.1\;\mathrm{mm} = 0.1\;\mathrm{mm}^3 = 10^{-4}\;\mathrm{mL}$

Total cells in the central square: $42 \times 25 = 1050$ cells

Cell concentration: $\frac{1050\;\mathrm{cells}}{10^{-4}\;\mathrm{mL}} = 1.05 \times 10^7\;\mathrm{cells/mL}$

This technique is essential in microbiology for determining cell density in culture, monitoring growth curves, and standardising inocula for experiments. A dilution factor must be applied if the suspension was diluted before loading.

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Common Pitfalls (Expanded)

• Confusing chromosomes and chromatids: a replicated chromosome consists of two sister chromatids joined at a centromere. After anaphase, each chromatid is an individual chromosome. Count chromosomes by counting centromeres, not arms.
• Equating osmosis with diffusion: osmosis refers specifically to water movement across a selectively permeable membrane. Diffusion is the general movement of particles from high to low concentration.
• Stating that prokaryotes "have no DNA": they possess a circular chromosome and plasmids, but lack a membrane-bound nucleus. The nucleoid region contains the DNA.
• Describing the cell membrane as a "solid" structure: it is fluid, with phospholipids and proteins able to move laterally (demonstrated by the Frye-Edidin fusion experiment, 1970).
• Confusing primary and secondary active transport: primary active transport directly hydrolyses ATP (e.g., the $\mathrm{Na}^+/\mathrm{K}^+$ pump). Secondary active transport uses the gradient established by primary transport as an energy source (e.g., $\mathrm{Na}^+$-glucose co-transport).
• Writing that mitosis produces "identical" cells without qualification: while genetically identical (barring mutation), the two daughter cells receive slightly different cytoplasmic contents, which can influence cell fate in early development.
• Assuming all eukaryotic cells have mitochondria: some anaerobic eukaryotes (e.g., certain parasitic protists) lack mitochondria entirely or possess reduced versions called mitosomes or hydrogenosomes.

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Exam-Style Problems

Problem 1: Data Analysis -- Osmosis Experiment

A student places potato cylinders of identical dimensions ($5\;\mathrm{cm} \times 0.5\;\mathrm{cm}$ diameter) into a series of sucrose solutions ranging from $0.0$ to $0.6\;\mathrm{mol/L}$ at $0.1\;\mathrm{mol/L}$ intervals. After $24$ hours at $20^\circ\mathrm{C}$, the student measures the mass change of each cylinder. The results are:

Sucrose concentration (mol/L)	Mass change (%)
0.0	+12.4
0.1	+7.8
0.2	+3.1
0.3	-1.2
0.4	-8.5
0.5	-14.3
0.6	-21.0

(a) Plot these data and estimate the water potential of the potato tissue. (b) Explain the biological basis for the trend observed. (c) Predict the result if the experiment were repeated at $35^\circ\mathrm{C}$.

Problem 2: Extended Response -- Endosymbiotic Theory

Mitochondria possess their own circular DNA, $70\mathrm{S}$ ribosomes, a double membrane, and the ability to replicate independently by binary fission. Evaluate the extent to which these features support the endosymbiotic theory. In your response, address at least two alternative explanations and discuss additional evidence (e.g., molecular phylogenetics) that strengthens the argument.

Problem 3: Quantitative -- Chi-Squared Test on Cell Cycle Data

A researcher observes $200$ cells under a microscope and records their mitotic phases:

Phase	Observed	Expected
Interphase	158	160
Prophase	22	16
Metaphase	10	8
Anaphase	6	8
Telophase	4	8

The expected values assume each mitotic phase occupies equal time. Use the chi-squared test ($\chi^2 = \sum \frac{(O - E)^2}{E}$) at the $p = 0.05$ significance level (critical value $= 9.49$ for $4$ degrees of freedom) to determine whether the observed distribution differs significantly from the expected. State your null hypothesis.

Problem 4: Diagram Analysis -- Comparing Prokaryotic and Eukaryotic Cells

An electron micrograph shows a cell approximately $2\;\mathrm{\mu m}$ in diameter with no internal membrane-bound organelles. Visible structures include a dense nucleoid region, free ribosomes, a capsule, and several pili. (a) Identify this cell as prokaryotic or eukaryotic and justify with three structural observations. (b) The cell is found in an environment with high antibiotic concentration. Explain why penicillin would be ineffective against this organism if it is an archaeon rather than a bacterium.

Problem 5: Extended Response -- Membrane Fluidity

Cholesterol modulates cell membrane fluidity. Describe and explain the dual role of cholesterol in membrane fluidity at both low and high temperatures. Discuss how this relates to the observation that cold-water fish species have a higher proportion of unsaturated fatty acids in their membrane phospholipids compared with warm-water species.

Problem 6: Quantitative -- Cell Division and Tumour Growth

A tumour starts from a single cell and undergoes mitosis every $24$ hours. Assuming $100\%$ cell survival and no apoptosis, calculate the number of cells after $14$ days. If the average cell diameter is $10\;\mathrm{\mu m}$ and the tumour grows as a sphere, estimate the tumour diameter after $14$ days. Explain why actual tumour growth is slower than this theoretical maximum.

Problem 7: Data Analysis -- Fick's Law Application

Gas exchange in the alveoli depends on the rate of diffusion described by Fick's Law. Alveolar surface area is approximately $70\;\mathrm{m}^2$ and the diffusion distance (alveolar wall + capillary wall) is approximately $1.5\;\mathrm{\mu m}$. The $\mathrm{O}_2$ partial pressure difference is approximately $8\;\mathrm{kPa}$. (a) Calculate the relative rate of $\mathrm{O}_2$ diffusion under these conditions. (b) Explain how emphysema, which destroys alveolar walls and reduces total surface area to $35\;\mathrm{m}^2$, would affect this rate.

Problem 8: Extended Response -- Specialised Cells and SA:V Ratio

Choose three specialised cells from different organ systems (e.g., small intestine epithelium, alveolar epithelium, proximal convoluted tubule). For each cell, describe one structural adaptation that increases surface area, and explain how this adaptation relates to the cell's physiological function. Use the concept of surface-area-to-volume ratio in your explanation.

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If You Get These Wrong, Revise:

• Mitosis and meiosis phases --> Review ./genetics
• DNA replication and chromosome structure --> Review ./molecular-biology
• Membrane transport and ATP use --> Review ./human-physiology
• Osmosis and water potential --> Review ./plant-biology
• Evolution and natural selection --> Review ./ecology

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8. Cell Specialisation and Stem Cells

Stem Cells

Definition. Stem cells are undifferentiated cells that have the capacity to self-renew (divide to produce more stem cells) and differentiate into specialised cell types.

Type	Source	Potency	Examples of differentiation
Totipotent	Zygote, early morula (up to 8-cell)	Can form any cell type, including extraembryonic tissues (placenta, yolk sac).	All cell types + embryonic membranes.
Pluripotent	Inner cell mass of blastocyst; embryonic stem cells (ESCs)	Can form any cell type of the body (all three germ layers), but NOT extraembryonic tissues.	Neurones, muscle, blood, liver, pancreatic cells, etc.
Multipotent	Adult tissues (bone marrow, skin, gut)	Can form multiple but limited cell types within a given lineage.	Haematopoietic stem cells $\to$ all blood cell types; mesenchymal stem cells $\to$ bone, cartilage, fat.
Unipotent	Specific adult tissues	Can form only one specialised cell type.	Spermatogonia $\to$ sperm; satellite cells $\to$ skeletal muscle.

Therapeutic Potential of Stem Cells

• Repair of damaged tissues: stem cell transplants for leukaemia (bone marrow transplants), spinal cord injury, Parkinson's disease (dopaminergic neurones), heart disease (cardiomyocytes).
• Drug testing: differentiated cells derived from stem cells can be used for pharmacological screening, reducing the need for animal testing.
• Disease modelling: induced pluripotent stem cells (iPSCs) from patients with genetic disorders allow researchers to study disease mechanisms in vitro.

Induced Pluripotent Stem Cells (iPSCs)

iPSCs are adult somatic cells that have been reprogrammed to a pluripotent state by the introduction of transcription factors (originally Oct4, Sox2, Klf4, c-Myc --- Yamanaka factors, 2006).

Advantages over embryonic stem cells: no ethical concerns about embryo destruction; patient-specific (matching the donor's genetics, reducing rejection risk).

Limitations: incomplete reprogramming (residual epigenetic memory); risk of tumour formation (tumorigenicity due to c-Myc oncogene); lower efficiency than ESCs.

Cell Cycle Regulation

The cell cycle is regulated at checkpoints by cyclins and cyclin-dependent kinases (CDKs):

Checkpoint	Location	Cyclin-CDK complex	Function
G1/S checkpoint	Late G1	Cyclin D-CDK4/6; Cyclin E-CDK2	Checks cell size, nutrient availability, DNA damage. If conditions are unfavourable, the cell enters G0 (quiescence). The retinoblastoma protein (Rb) is a key regulator: when hypophosphorylated, Rb binds E2F (transcription factor) and blocks S-phase gene expression; cyclin D-CDK4/6 phosphorylates Rb, releasing E2F.
G2/M checkpoint	Late G2	Cyclin B-CDK1 (CDC2)	Checks that DNA replication is complete and no DNA damage remains. Activated by phosphorylation by CDK-activating kinase (CAK).
Spindle assembly checkpoint (M checkpoint)	Metaphase	Mad2, BubR1 proteins	Ensures all chromosomes are correctly attached to spindle fibres via kinetochores before anaphase begins. Unattached kinetochores generate a "wait" signal that inhibits the anaphase-promoting complex/cyclosome (APC/C).

Apoptosis (Programmed Cell Death)

Apoptosis is a tightly regulated process of cell death that eliminates damaged, infected, or unwanted cells without triggering inflammation.

Characteristics:

• Cell shrinks; chromatin condenses (pyknosis).
• DNA is fragmented into regular-sized pieces (ladder pattern on gel electrophoresis).
• The cell breaks into apoptotic bodies, which are phagocytosed by macrophages.
• Cell membrane remains intact (no release of intracellular contents, no inflammation).

Mechanism:

1. Extrinsic pathway: death receptors (e.g., Fas/CD95) on the cell surface bind to death ligands (e.g., FasL on cytotoxic T cells), activating caspase-8.
2. Intrinsic (mitochondrial) pathway: cellular stress (DNA damage, oxidative stress) triggers pro-apoptotic Bcl-2 family proteins (Bax, Bak) to permeabilise the mitochondrial outer membrane, releasing cytochrome c. Cytochrome c binds to Apaf-1, forming the apoptosome, which activates caspase-9.
3. Execution phase: initiator caspases (8, 9) activate effector caspases (3, 7), which systematically degrade cellular proteins, including PARP and lamins.

Anti-apoptotic proteins (Bcl-2, Bcl-xL) inhibit the intrinsic pathway. Overexpression of Bcl-2 is associated with some cancers (cells fail to undergo apoptosis).

Cancer and the Cell Cycle

Cancer is a disease of uncontrolled cell division resulting from mutations in genes that regulate the cell cycle.

Gene type	Normal function	Mutated in cancer (oncogene / tumour suppressor)
Proto-oncogenes	Promote cell division	Oncogenes: gain-of-function mutations; hyperactive or overexpressed (e.g., ras, HER2/neu, myc).
Tumour suppressor genes	Inhibit cell division; promote DNA repair; promote apoptosis	Loss-of-function mutations (e.g., p53, Rb, BRCA1, BRCA2). Both alleles must be mutated (Knudson's two-hit hypothesis).

p53 ("guardian of the genome"):

• Activated by DNA damage (detected by ATM/ATR kinases).
• Can induce cell cycle arrest (via p21, a CDK inhibitor) to allow DNA repair.
• Can induce apoptosis (via Bax, PUMA) if damage is irreparable.
• Mutated in $\approx 50\%$ of all human cancers, making it the most commonly mutated gene in cancer.

Metastasis: the spread of cancer cells from the primary tumour to secondary sites via the bloodstream or lymphatic system. Requires: local invasion, intravasation, survival in circulation, extravasation, and colonisation of distant tissues.

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9. Prokaryotic Cell Biology (Extended)

Bacterial Cell Wall Structure

The peptidoglycan (murein) layer is a polymer of alternating $N$-acetylglucosamine (NAG) and $N$-acetylmuramic acid (NAM) sugars, cross-linked by short peptide bridges (containing D-alanine, D-glutamate, and a diaminopimelic acid-lysine cross-link).

Gram-positive bacteria: thick peptidoglycan layer ($20$--$80\;\mathrm{nm}$) with teichoic acids embedded. Retain crystal violet stain in Gram staining. Gram-negative bacteria: thin peptidoglycan layer ($2$--$7\;\mathrm{nm}$) with an outer membrane containing lipopolysaccharide (LPS). Do not retain crystal violet; appear pink with safranin counterstain.

Antibiotic targets in the cell wall:

• Penicillin: inhibits transpeptidase (penicillin-binding proteins), preventing peptide cross-linking. Effective against Gram-positive bacteria; Gram-negative bacteria are resistant due to their outer membrane (though some penicillins can penetrate).
• Lysozyme: hydrolyses the $\beta$-1,4-glycosidic bond between NAM and NAG; found in tears, saliva, and egg white.

Bacterial Growth Curve

When bacteria are inoculated into fresh medium:

Phase	Description
Lag phase	No immediate increase in cell number; cells are adapting, synthesising enzymes, and taking up nutrients.
Exponential (log) phase	Cells divide at maximum rate (binary fission); population doubles at a constant rate. $N = N_0 \times 2^n$
Stationary phase	Growth rate = death rate; nutrients are depleted, waste products accumulate, and space is limited.
Decline (death) phase	Death rate exceeds growth rate; cells die from nutrient starvation and toxin accumulation.

Growth rate: $\mu = \frac{\ln(N_t / N_0)}{t}$ (specific growth rate, $\mathrm{h}^{-1}$).

Doubling time: $t_d = \frac{\ln 2}{\mu} = \frac{0.693}{\mu}$

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10. Advanced Membrane Biology

Membrane Fluidity

The fluidity of the cell membrane depends on:

• Fatty acid composition: more unsaturated (C=C) fatty acids increase fluidity (kinks prevent tight packing); more saturated fatty acids decrease fluidity.
• Cholesterol content: at high temperatures, cholesterol restrains movement (decreases fluidity); at low temperatures, it prevents tight packing (increases fluidity). Acts as a fluidity buffer.
• Temperature: higher temperatures increase kinetic energy and fluidity; lower temperatures decrease it.

Membrane Proteins: Types and Functions

Integral (transmembrane) proteins:

• Channel proteins: form hydrophilic pores for specific ions ($\mathrm{Na}^+$, $\mathrm{K}^+$, $\mathrm{Ca}^{2+}$, $\mathrm{Cl}^-$). Can be gated (voltage-gated, ligand-gated, mechanically-gated) or ungated (leak channels).
• Carrier proteins (transporters): undergo conformational change to shuttle substances across. Include uniporters, symporters (co-transport in same direction), and antiporters (co-transport in opposite direction).
• Receptors: transmembrane proteins that bind signalling molecules (hormones, neurotransmitters) and initiate intracellular signal transduction cascades.

Peripheral proteins:

• Attached to the cytoplasmic or extracellular face via electrostatic interactions with integral proteins or lipid anchors.
• Include cytoskeletal anchoring proteins (spectrin, ankyrin in erythrocytes) and enzymes (adenylate cyclase, phospholipase C).

Signal Transduction Across Membranes

Hydrophilic signalling molecules (peptide hormones, neurotransmitters) cannot cross the lipid bilayer and must bind to cell-surface receptors:

1. G-protein coupled receptors (GPCRs): the ligand binds to the receptor, causing a conformational change that activates a G-protein (G$\alpha$, G$\beta$, G$\gamma$ subunits). The G$\alpha$ subunit exchanges GDP for GTP and activates downstream effectors (adenylate cyclase $\to$ cAMP; phospholipase C $\to$ IP$_3$ + DAG).
2. Receptor tyrosine kinases (RTKs): ligand binding causes receptor dimerisation and autophosphorylation of tyrosine residues, creating docking sites for intracellular signalling proteins (e.g., the MAP kinase cascade).

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11. Cell Fractionation and Microscopy Techniques

Cell Fractionation

The process of breaking open cells and separating organelles by differential centrifugation:

1. Homogenisation: cells are broken open in an isotonic, buffered solution (to preserve organelle structure) using a homogeniser or sonication.
2. Differential centrifugation:
   • Low speed ($1000\;\mathrm{g}$, $10\;\mathrm{min}$): nuclei and unbroken cells pellet.
   • Medium speed ($10000\;\mathrm{g}$, $20\;\mathrm{min}$): mitochondria, lysosomes, peroxisomes pellet.
   • High speed ($100000\;\mathrm{g}$, $60\;\mathrm{min}$): ribosomes, microsomes (ER fragments) pellet.
   • Supernatant: cytosol (soluble proteins and enzymes).

Microscopy Techniques

Technique	Resolution	Sample preparation	Information obtained
Light microscopy	$\approx 200\;\mathrm{nm}$	Staining; live specimens possible	Overall cell structure; limited detail of organelles.
Electron microscopy (TEM)	$\approx 0.5\;\mathrm{nm}$	Fixed, dehydrated, heavy metal staining; thin sections	Ultrastructure of organelles; internal detail.
Electron microscopy (SEM)	$\approx 3$--$20\;\mathrm{nm}$	Fixed, dehydrated, metal-coated; surface view	3D surface topography of cells and organisms.
Fluorescence microscopy	$\approx 200\;\mathrm{nm}$	Fluorescent dyes or fluorescent proteins (GFP)	Location of specific molecules in living cells.
Confocal microscopy	$\approx 200\;\mathrm{nm}$	Same as fluorescence	Optical sections; 3D reconstruction; reduced blur.
Cryo-electron microscopy	$\approx 3\;\mathrm{nm}$	Flash-frozen, no staining	Near-native state structure of large complexes.

Magnification calculation:

$\mathrm{Magnification} = \frac{\text{image size}}{\text{actual size}}$

Resolution: the minimum distance between two points that can be distinguished as separate. The resolving power of a microscope is limited by the wavelength of the illuminating radiation:

$d = \frac{0.61\lambda}{n \sin\theta}$

where $\lambda$ is the wavelength, $n$ is the refractive index of the medium, and $\theta$ is the half-angle of the cone of light entering the objective.

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Exam-Style Problems (Extended)

Problem 9: Extended Response -- Stem Cells and Ethics

Compare and contrast embryonic stem cells (ESCs) and induced pluripotent stem cells (iPSCs) with regard to: (a) their origin and method of production, (b) their potency, (c) their therapeutic potential, and (d) the ethical issues associated with each. Discuss the significance of the Yamanaka factors in the development of iPSC technology.

Problem 10: Data Analysis -- Cell Cycle and Mitotic Index

A researcher counts the number of cells in each stage of mitosis in a sample of onion root tip cells. The results are: interphase = 800, prophase = 60, metaphase = 30, anaphase = 10, telophase = 10. Total cells counted = 910. (a) Calculate the mitotic index (percentage of cells undergoing mitosis). (b) Estimate the duration of mitosis if the cell cycle takes 20 hours. (c) Explain why the majority of cells are in interphase. (d) If colchicine (which disrupts spindle fibre formation) is applied to the root tip, predict the effect on the mitotic index and explain why.

Problem 11: Extended Response -- Cancer Biology

BRCA1 is a tumour suppressor gene involved in DNA repair. Inherited mutations in BRCA1 increase the risk of breast and ovarian cancer. (a) Explain the normal function of BRCA1 in the cell cycle. (b) Using Knudson's two-hit hypothesis, explain why individuals with one inherited BRCA1 mutation have a higher cancer risk than the general population. (c) Explain why a double mastectomy reduces cancer risk in BRCA1 carriers. (d) Discuss the ethical considerations of genetic testing for BRCA1 mutations.

Problem 12: Quantitative -- Surface Area and Diffusion

A spherical human egg cell has a diameter of $120\;\mathrm{\mu m}$. (a) Calculate its surface area and volume. (b) After fertilisation, the zygote undergoes 6 rounds of cleavage (mitosis without growth) to form a morula of $64$ cells. Assuming the morula is roughly spherical with the same total volume as the original zygote, calculate the surface-area-to-volume ratio of each cell in the morula (assuming equal cell sizes). (c) Explain the biological significance of the change in SA:V ratio during early embryonic development.

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Additional Worked Examples

Worked Example: Prokaryotic vs Eukaryotic Cell Comparison

(a) Calculate the volume of a typical E. coli cell (rod-shaped, $2\;\mathrm{\mu m}$ long, $0.5\;\mathrm{\mu m}$ diameter) and a typical human liver cell (spherical, $20\;\mathrm{\mu m}$ diameter). (b) Calculate the surface-area-to-volume ratio for each. (c) Explain why prokaryotes can rely on simple diffusion for many processes while eukaryotes require specialised transport systems.

Solution

(a) E. coli (approximate as cylinder with hemispherical ends): Volume of cylindrical part: $\pi r^2 h = \pi \times (0.25)^2 \times 1.5 = 0.295\;\mathrm{\mu m}^3$ Volume of hemispherical ends: $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.25)^3 = 0.065\;\mathrm{\mu m}^3$ Total volume: $0.295 + 0.065 = 0.36\;\mathrm{\mu m}^3$

Human liver cell (sphere): Volume $= \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(10)^3 = 4189\;\mathrm{\mu m}^3$

(b) E. coli surface area: Cylindrical part: $2\pi rh = 2\pi \times 0.25 \times 1.5 = 2.36\;\mathrm{\mu m}^2$ Hemispherical ends: $4\pi r^2 = 4\pi(0.25)^2 = 0.785\;\mathrm{\mu m}^2$ Total SA: $2.36 + 0.785 = 3.14\;\mathrm{\mu m}^2$ SA:V $= 3.14 / 0.36 = 8.7\;\mathrm{\mu m}^{-1}$

Human liver cell SA $= 4\pi r^2 = 4\pi(10)^2 = 1257\;\mathrm{\mu m}^2$ SA:V $= 1257 / 4189 = 0.30\;\mathrm{\mu m}^{-1}$

(c) E. coli has a SA:V ratio approximately $29\times$ higher than the liver cell. This high ratio means that the cell surface is large relative to its volume, allowing efficient diffusion of oxygen, nutrients, and waste products throughout the cell. The small size also means that no internal structure is far from the cell membrane. Eukaryotic cells, being much larger, cannot rely on diffusion alone; they require: endomembrane systems (ER, Golgi) for intracellular transport, cytoskeletal tracks for vesicle movement, specialised organelles (mitochondria for ATP production, lysosomes for waste processing), and membrane transport proteins for selective uptake and export.

Worked Example: Osmosis and Water Potential

A plant cell with a water potential of $-400\;\mathrm{kPa}$ is placed in a solution with water potential of $-200\;\mathrm{kPa}$. The pressure potential of the cell is $+100\;\mathrm{kPa}$. (a) Calculate the solute potential of the cell. (b) Predict the direction of water movement. (c) Describe what happens to the cell over time (include changes in turgor pressure, volume, and water potential). (d) Repeat the analysis if the external solution has a water potential of $-800\;\mathrm{kPa}$.

Solution

(a) $\Psi = \Psi_s + \Psi_p$, so $\Psi_s = \Psi - \Psi_p = -400 - 100 = -500\;\mathrm{kPa}$.

(b) Water moves from higher water potential to lower water potential. External $= -200\;\mathrm{kPa}$; cell $= -400\;\mathrm{kPa}$. Water moves into the cell (from $-200$ to $-400\;\mathrm{kPa}$).

(c) As water enters:

• The cell volume increases (the cell wall restricts expansion, so the cell becomes turgid).
• Turgor pressure ($\Psi_p$) increases.
• As $\Psi_p$ increases, the cell's water potential ($\Psi = \Psi_s + \Psi_p$) becomes less negative.
• Equilibrium is reached when $\Psi_{cell} = \Psi_{external} = -200\;\mathrm{kPa}$: $-500 + \Psi_p = -200$, so $\Psi_p = +300\;\mathrm{kPa}$.
• The cell is fully turgid.

(d) External $= -800\;\mathrm{kPa}$; cell $= -400\;\mathrm{kPa}$. Water moves out of the cell.

• The cell volume decreases.
• Turgor pressure decreases (from $+100$ toward $0$).
• Eventually, the cell membrane pulls away from the cell wall (plasmolysis).
• $\Psi_p$ reaches $0$ (no pressure against the wall). The cell's water potential $= \Psi_s = -500\;\mathrm{kPa}$.
• Since $-500 > -800$, water continues to leave until the solute potential becomes more negative (concentration increases). Equilibrium when $\Psi_{cell} = -800\;\mathrm{kPa}$: $\Psi_s + 0 = -800\;\mathrm{kPa}$, so $\Psi_s = -800\;\mathrm{kPa}$. The cell has lost significant water and is plasmolysed.

Worked Example: Membrane Transport Quantification

A red blood cell has an internal $\mathrm{K}^+$ concentration of $140\;\mathrm{mmol/L}$ and an external $\mathrm{K}^+$ concentration of $5\;\mathrm{mmol/L}$ at $37^\circ\mathrm{C}$. The membrane potential is $-12\;\mathrm{mV}$ (inside negative). (a) Calculate the $\mathrm{K}^+$ equilibrium potential. (b) Calculate the net electrochemical driving force on $\mathrm{K}^+$. (c) In which direction does the $\mathrm{Na}^+/K^+$ pump move $\mathrm{K}^+$? (d) How many ATP molecules are needed to pump $1000\;\mathrm{K}^+$ ions into the cell?

Solution

(a) $E_{\mathrm{K}} = \frac{RT}{zF}\ln\frac{[\mathrm{K}^+]_{out}}{[\mathrm{K}^+]_{in}} = \frac{8.314 \times 310}{1 \times 96485}\ln\frac{5}{140}$ $= 0.0267 \times \ln(0.0357) = 0.0267 \times (-3.332) = -88.9\;\mathrm{mV}$

(b) Net driving force $= V_m - E_{\mathrm{K}} = -12 - (-88.9) = +76.9\;\mathrm{mV}$. The positive value means the driving force is outward (favouring $\mathrm{K}^+$ efflux).

(c) The $\mathrm{Na}^+/K^+$ pump moves $\mathrm{K}^+$ into the cell (against its concentration gradient and electrochemical gradient), using ATP to actively transport $\mathrm{K}^+$ from the extracellular fluid ($5\;\mathrm{mmol/L}$) to the intracellular fluid ($140\;\mathrm{mmol/L}$).

(d) The $\mathrm{Na}^+/K^+$ pump moves $2\;\mathrm{K}^+$ in per ATP hydrolysed. ATP needed $= 1000 / 2 = 500$ ATP molecules.

In molar terms: $1000\;\mathrm{K}^+$ ions $= 1000 / (6.022 \times 10^{23}) = 1.66 \times 10^{-21}\;\mathrm{mol\;K}^+$. ATP $= 500 / (6.022 \times 10^{23}) = 8.30 \times 10^{-22}\;\mathrm{mol\;ATP}$.

Worked Example: Cell Cycle and Mitotic Index

A researcher counts cells in an onion root tip preparation:

• Interphase: $250$ cells
• Prophase: $30$ cells
• Metaphase: $15$ cells
• Anaphase: $10$ cells
• Telophase: $5$ cells Total: $310$ cells

(a) Calculate the mitotic index. (b) If the total cell cycle takes $20$ hours, estimate the duration of each phase. (c) Explain why the mitotic index varies between different tissues. (d) A mutagen is applied to the root tip, and the mitotic index drops to $5\%$. Explain two possible mechanisms.

Solution

(a) Mitotic index $= \frac{\text{cells in mitosis}}{\text{total cells}} \times 100$ $= \frac{30 + 15 + 10 + 5}{310} \times 100 = \frac{60}{310} \times 100 = 19.4\%$

(b) Duration of each phase is proportional to the fraction of cells in that phase: Total time $= 20\;\mathrm{h} = 1200\;\mathrm{min}$.

• Interphase: $\frac{250}{310} \times 1200 = 968\;\mathrm{min} = 16.1\;\mathrm{h}$
• Prophase: $\frac{30}{310} \times 1200 = 116\;\mathrm{min} = 1.9\;\mathrm{h}$
• Metaphase: $\frac{15}{310} \times 1200 = 58\;\mathrm{min}$
• Anaphase: $\frac{10}{310} \times 1200 = 39\;\mathrm{min}$
• Telophase: $\frac{5}{310} \times 1200 = 19\;\mathrm{min}$

(c) The mitotic index varies between tissues because different tissues have different rates of cell division:

• Rapidly dividing tissues (root tips, bone marrow, skin basal layer, intestinal crypts) have high mitotic indices ($10$--$20\%$).
• Slowly dividing or non-dividing tissues (neurons, cardiac muscle, mature lymphocytes) have very low mitotic indices ($<1\%$).
• The mitotic index reflects the proportion of time cells spend in mitosis relative to the total cell cycle, which depends on the tissue's need for cell renewal.

(d) Possible mechanisms for a reduced mitotic index after mutagen exposure:

1. Cell cycle arrest: DNA damage activates checkpoint proteins (p53, ATM/ATR) that arrest the cell cycle at G1/S or G2/M checkpoints, preventing cells from entering mitosis until DNA is repaired.
2. Apoptosis: if DNA damage is too severe to repair, the cell undergoes programmed cell death, reducing the total number of cells and the number of cells entering mitosis.
3. Cell death (necrosis): the mutagen may be cytotoxic, directly killing cells.
4. Senescence: damaged cells may enter a permanent state of cell cycle arrest (senescence), no longer dividing.

Worked Example: Electron Microscopy and Resolution

(a) Calculate the resolving power of a light microscope with a numerical aperture (NA) of $1.4$ using green light ($\lambda = 550\;\mathrm{nm}$). (b) Calculate the resolving power of a transmission electron microscope (TEM) with an accelerating voltage of $80\;\mathrm{kV}$ ($\lambda \approx 0.005\;\mathrm{nm}$) and NA $= 0.02$ (based on aperture angle). (c) A ribosome is approximately $25\;\mathrm{nm}$ in diameter. Can it be resolved with each microscope? (d) Explain the difference between TEM and SEM in terms of image formation and the type of specimens each is best suited for.

Solution

(a) Resolution $d = \frac{0.61\lambda}{\mathrm{NA}} = \frac{0.61 \times 550}{1.4} = 240\;\mathrm{nm} = 0.24\;\mathrm{\mu m}$.

(b) TEM resolution $d = \frac{0.61 \times 0.005}{0.02} = 0.153\;\mathrm{nm}$.

(In practice, TEM resolution is limited by lens aberrations to approximately $0.1$--$0.2\;\mathrm{nm}$, which is sufficient to resolve individual atoms in crystalline specimens.)

(c) Ribosome diameter $= 25\;\mathrm{nm}$.

• Light microscope: resolution $= 240\;\mathrm{nm}$. $25 < 240$, so the ribosome is smaller than the resolution limit. It cannot be resolved (individual ribosomes are invisible under light microscopy).
• TEM: resolution $= 0.153\;\mathrm{nm}$. $25 > 0.153$, so the ribosome is easily resolved. TEM can clearly visualise ribosomes and their subunits.

(d) TEM: a beam of electrons passes through an ultra-thin specimen ($<100\;\mathrm{nm}$ thick). Denser regions scatter more electrons, appearing dark in the image (bright-field imaging). TEM produces 2D images showing internal cellular ultrastructure (organelles, membranes, ribosomes, cytoskeleton). Specimens must be fixed, dehydrated, and embedded in resin; staining with heavy metals (uranyl acetate, lead citrate) enhances contrast.

SEM: a focused beam of electrons scans the surface of a specimen. Secondary electrons emitted from the surface are collected to produce a 3D-like image of the surface topography. SEM shows external surface features (cell surface, organelle surfaces, microvilli) with great depth of field. Specimens are coated with a thin layer of metal (gold or platinum) to enhance electron emission.

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Additional Common Pitfalls

• Confusing resolution and magnification: magnification makes things bigger; resolution determines whether two closely spaced objects can be distinguished as separate. Increasing magnification beyond the resolution limit produces a larger but blurry image.
• Stating that prokaryotes have no internal membranes: many prokaryotes have internal membrane infoldings (mesosomes, thylakoids in cyanobacteria, chromatophores in photosynthetic bacteria).
• Confusing osmosis and diffusion: osmosis is the specific case of diffusion of water across a selectively permeable membrane; diffusion is the general movement of molecules from high to low concentration.
• Assuming all active transport uses ATP directly: some secondary active transport uses the electrochemical gradient established by a primary active transporter (e.g., $\mathrm{Na}^+$/glucose co-transport uses the $\mathrm{Na}^+$ gradient created by the $\mathrm{Na}^+/K^+$ pump).
• Confusing chromosomes and chromatids: a chromosome consists of one or two chromatids. Before replication: 1 chromosome = 1 chromatid. After replication: 1 chromosome = 2 sister chromatids. After anaphase: each chromatid is now an independent chromosome.
• Forgetting that mitosis produces genetically identical daughter cells: the purpose of mitosis is growth, repair, and asexual reproduction; meiosis produces genetically unique gametes.

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Additional Exam-Style Problems with Full Solutions

Problem 13: Extended Response -- Endosymbiotic Theory

(a) Describe the endosymbiotic theory for the origin of mitochondria and chloroplasts. (b) Provide four pieces of evidence that support this theory. (c) Explain why mitochondria and chloroplasts have retained some of their own genes despite having transferred most genes to the nucleus. (d) Explain why mitochondria are inherited maternally in most animals.

Answer 13

(a) The endosymbiotic theory proposes that mitochondria and chloroplasts originated from free-living prokaryotes that were engulfed by a larger host cell (probably an archaeon). The engulfed prokaryotes formed a symbiotic relationship with the host: the host provided protection and nutrients; the endosymbionts provided ATP (mitochondria, from an alpha-proteobacterium) or organic compounds from photosynthesis (chloroplasts, from a cyanobacterium). Over billions of years, the endosymbionts lost many genes (transferred to the host nucleus) and became integrated as organelles.

(b) Evidence:

1. Double membrane: mitochondria and chloroplasts are surrounded by two membranes; the inner membrane is thought to be the original prokaryotic membrane, and the outer membrane is derived from the host's phagocytic vacuole.
2. Own DNA: both organelles contain circular DNA, similar to prokaryotic chromosomes (not linear like nuclear DNA). The DNA does not associate with histones.
3. 70S ribosomes: both contain ribosomes of similar size and structure to prokaryotic ribosomes (70S, not 80S like eukaryotic cytoplasmic ribosomes). They are sensitive to antibiotics that inhibit prokaryotic protein synthesis (e.g., chloramphenicol, tetracycline).
4. Binary fission: both reproduce by binary fission, not by mitosis.
5. Phylogenetic evidence: ribosomal RNA sequences from mitochondria are most closely related to alpha-proteobacteria; chloroplast rRNA is most closely related to cyanobacteria.
6. Transit peptides: proteins imported into mitochondria and chloroplasts have N-terminal signal sequences that are cleaved after import, similar to prokaryotic signal peptides.

(c) Mitochondria and chloroplasts have retained some genes because:

• The encoded proteins are highly hydrophobic (e.g., membrane proteins of the electron transport chain) and would be difficult to import through the organelle membranes if encoded in the nucleus.
• Local control of gene expression allows rapid response to changes in the organelle's environment (e.g., redox state) without waiting for nuclear gene regulation.
• Co-translational insertion of hydrophobic proteins into the inner membrane requires local mRNA translation by organelle ribosomes.

(d) Maternal inheritance of mitochondria occurs because:

• The sperm contributes very few (if any) mitochondria to the zygote. The mitochondria in the sperm's midpiece are destroyed after fertilisation (in many species, the sperm mitochondria are targeted for degradation by ubiquitin-proteasome pathways in the oocyte).
• The egg (oocyte) contains a large number of mitochondria ($>100\,000$), which vastly outnumber any sperm-derived mitochondria.
• This means all mitochondrial DNA in an individual is derived from the mother, which is the basis for mitochondrial DNA tracing in evolutionary and forensic studies (mitochondrial Eve).

Problem 14: Quantitative -- Serial Dilution and Cell Counting

A bacterial culture has an OD600 (optical density at 600 nm) of 1.2. A calibration curve shows that OD600 = 1.0 corresponds to $8 \times 10^8$ cells/mL. A serial dilution is performed: $100\;\mathrm{\mu L}$ of culture is added to $900\;\mathrm{\mu L}$ of sterile broth (dilution factor $10^{-1}$), and this is repeated for 6 serial dilutions. $100\;\mathrm{\mu L}$ of the $10^{-6}$ dilution is spread on an agar plate, yielding $72$ colonies. (a) Calculate the cell concentration in the original culture from the OD600. (b) Calculate the viable cell concentration from the plate count. (c) Calculate the percentage of viable cells. (d) Explain why the OD600 count and the viable count differ.

Answer 14

(a) From OD600: concentration $= 1.2 \times 8 \times 10^8 = 9.6 \times 10^8$ cells/mL.

(b) From plate count: $72$ colonies on the $10^{-6}$ dilution plate, with $100\;\mathrm{\mu L}$ plated. Concentration $= \frac{72}{0.1\;\mathrm{mL}} \times 10^6 = 7.2 \times 10^8$ cells/mL.

(c) Percentage viable $= \frac{7.2 \times 10^8}{9.6 \times 10^8} \times 100 = 75\%$.

(d) The OD600 counts all cells (both viable and dead) because it measures light scattering by all particles in suspension. The plate count only measures viable (living) cells that can form colonies. The $25\%$ difference represents dead or non-viable cells in the culture, which may have been killed by nutrient depletion, toxin accumulation, or being in a dormant (non-culturable) state.

Problem 15: Extended Response -- Stem Cells and Cell Differentiation

(a) Define totipotent, pluripotent, and multipotent stem cells, giving an example of each. (b) Describe the process of cell differentiation, explaining the role of transcription factors. (c) Explain how therapeutic cloning (somatic cell nuclear transfer) could be used to generate patient-specific stem cells. (d) Evaluate the ethical arguments for and against embryonic stem cell research.

Answer 15

(a) Totipotent: can differentiate into any cell type AND extraembryonic tissues (placenta, yolk sac). Example: the zygote and early blastomeres (up to the 4-cell stage in humans).

Pluripotent: can differentiate into any cell type of the body (all three germ layers: ectoderm, mesoderm, endoderm) but NOT extraembryonic tissues. Example: embryonic stem cells (ESCs) derived from the inner cell mass of the blastocyst.

Multipotent: can differentiate into a limited range of cell types within a particular lineage. Example: haematopoietic stem cells (bone marrow) can produce all blood cell types but not nerve or muscle cells; mesenchymal stem cells can produce bone, cartilage, and fat cells.

(b) Cell differentiation is the process by which unspecialised stem cells become specialised cell types with specific structures and functions. The fundamental mechanism is differential gene expression: all cells in an organism have the same genome, but different cell types express different subsets of genes. This is controlled by:

1. Transcription factors: proteins that bind to regulatory sequences (enhancers, promoters) of specific genes, activating or repressing their transcription. Combinations of transcription factors establish cell-type-specific gene expression programmes.
2. Epigenetic modifications: DNA methylation and histone modifications create heritable patterns of gene activation/silencing that are maintained through cell division (see ./genetics-advanced).
3. External signals: growth factors, hormones, and cell-cell interactions (e.g., Notch signalling) activate intracellular signalling cascades that modulate transcription factor activity.

Once a cell is committed to a particular lineage, its fate is progressively restricted by the activation of new transcription factors and the silencing of genes for alternative lineages.

(c) Therapeutic cloning (SCNT):

1. A somatic cell (e.g., skin fibroblast) is taken from the patient.
2. The nucleus is extracted from the somatic cell.
3. The nucleus is injected into an enucleated oocyte (egg cell with its nucleus removed).
4. The reconstructed oocyte is stimulated (electrically or chemically) to begin dividing.
5. After 5--7 days, a blastocyst forms. The inner cell mass is extracted to generate embryonic stem cells that are genetically identical to the patient.
6. These stem cells can be directed to differentiate into the required cell type (e.g., dopaminergic neurons for Parkinson's disease, insulin-producing beta cells for diabetes) and transplanted back into the patient without risk of immune rejection.

(d) Ethical arguments:

• For: potential to treat currently incurable diseases (Parkinson's, diabetes, spinal cord injury); the embryos used are typically surplus IVF embryos that would otherwise be destroyed; the potential to alleviate suffering outweighs the moral status of a blastocyst (a collection of approximately $150$ cells that has no nervous system, consciousness, or capacity for suffering).
• Against: the embryo has moral status from the moment of conception and should not be destroyed for research; creating embryos specifically for research commodifies human life; adult stem cells and induced pluripotent stem cells (iPSCs) offer alternatives that do not require embryo destruction; slippery slope argument: acceptance of embryonic research could lead to reproductive cloning.

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Cross-References to Related Topics

• Membrane transport and proteins: Review ./molecular-biology for membrane protein structure and fluid mosaic model.
• Enzymes and metabolism: Review ./metabolism-cell-biology for ATP, respiration, and photosynthesis in organelles.
• Cell division and meiosis: Review ./genetics for meiosis and genetic recombination.
• DNA replication in organelles: Review ./genetics-advanced for mitochondrial DNA replication and inheritance.
• Immune cells and phagocytosis: Review ./immunology for specialised cell types of the immune system.

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Supplementary: Membrane Structure and Transport in Detail (HL Extension)

The Fluid Mosaic Model

The fluid mosaic model (Singer and Nicolson, 1972) describes the cell membrane as a two-dimensional fluid of lipids and proteins. Key features:

Fluidity: phospholipids and proteins can move laterally within the membrane (lateral diffusion) but rarely flip between leaflets (transverse diffusion, or "flip-flop," requires flippase enzymes). Membrane fluidity depends on:

• Temperature: higher temperature = more fluid.
• Fatty acid composition: more unsaturated fatty acids (C=C double bonds create kinks) = more fluid.
• Cholesterol: at high temperatures, cholesterol restricts movement (reducing fluidity); at low temperatures, it prevents tight packing (maintaining fluidity). Cholesterol acts as a "fluidity buffer."

Asymmetry: the two leaflets of the membrane are not identical. For example:

• Glycolipids and glycoproteins are found on the extracellular face (cell recognition, signalling).
• Phosphatidylinositol is enriched on the cytoplasmic face (serves as a docking site for signalling proteins).
• Flippases, floppases, and scramblases maintain and regulate membrane asymmetry.

Types of Membrane Proteins

Integral (transmembrane) proteins: span the entire membrane, with hydrophobic regions (typically alpha helices) embedded in the lipid bilayer. Examples: ion channels, G-protein coupled receptors (GPCRs), transport proteins.

Peripheral proteins: associated with the membrane surface (extracellular or cytoplasmic face) via non-covalent interactions (electrostatic, hydrogen bonds) with integral proteins or lipid head groups. Examples: spectrin (cytoskeleton linkage), some enzymes.

Anchored proteins: attached to the membrane via a lipid modification (e.g., GPI anchor, prenylation, myristoylation). Examples: alkaline phosphatase (GPI-anchored), Ras (prenylated).

Types of Membrane Transport

Type	Direction	Energy	Carrier	Example
Simple diffusion	High to low conc.	None (passive)	None	$\mathrm{O}_2$, $\mathrm{CO}_2$, ethanol, steroid hormones
Facilitated diffusion	High to low conc.	None (passive)	Channel or carrier protein	Glucose (via GLUT transporters), $\mathrm{Na}^+$ (via leak channels)
Primary active transport	Low to high conc.	ATP	Pump protein	$\mathrm{Na}^+/K^+$ pump, $\mathrm{Ca}^{2+}$ pump, $\mathrm{H}^+/K^+$ pump
Secondary active transport	Low to high conc.	Ion gradient	Symporter or antiporter	$\mathrm{Na}^+$/glucose symporter (SGLT1), $\mathrm{Na}^+/\mathrm{Ca}^{2+}$ exchanger
Bulk transport	Variable	ATP	Vesicles	Endocytosis, exocytosis, phagocytosis

Channel Proteins

Ion channels are transmembrane pores that allow specific ions to pass through. They are gated (opened/closed) by specific stimuli:

• Voltage-gated channels: open in response to changes in membrane potential. Examples: voltage-gated $\mathrm{Na}^+$ channels (action potential initiation), voltage-gated $\mathrm{K}^+$ channels (action potential repolarisation), voltage-gated $\mathrm{Ca}^{2+}$ channels (neurotransmitter release).
• Ligand-gated channels: open when a specific ligand (neurotransmitter, hormone) binds. Examples: nicotinic ACh receptor (opens when ACh binds at NMJ), GABA_A receptor (opens when GABA binds, allowing $\mathrm{Cl}^-$ influx, hyperpolarising the cell).
• Mechanically-gated channels: open in response to physical deformation. Examples: stretch receptors in the skin, auditory hair cells in the cochlea.
• Aquaporins: channels specifically for water (not ions). Essential for rapid water transport in kidney collecting ducts (regulated by ADH) and plant cells.

The Sodium-Potassium Pump ($\mathrm{Na}^+/K^+$-ATPase)

The $\mathrm{Na}^+/K^+$ pump maintains the resting membrane potential and electrochemical gradients essential for nerve impulse transmission, secondary active transport, and osmotic balance.

Mechanism:

1. Three $\mathrm{Na}^+$ ions bind from the cytoplasmic side.
2. ATP is hydrolysed, transferring a phosphate group to the pump (phosphorylation).
3. Phosphorylation causes a conformational change, exposing the $\mathrm{Na}^+$ binding sites to the extracellular side. $\mathrm{Na}^+$ is released.
4. Two $\mathrm{K}^+$ ions bind from the extracellular side.
5. Dephosphorylation causes another conformational change, exposing $\mathrm{K}^+$ to the cytoplasmic side. $\mathrm{K}^+$ is released.

Net effect: $3\;\mathrm{Na}^+$ out, $2\;\mathrm{K}^+$ in, per ATP hydrolysed. Electrogenic: moves 1 net positive charge out per cycle (3 out, 2 in), contributing to the negative resting membrane potential (approximately $-3$ to $-5\;\mathrm{mV}$).

Energy cost: the $\mathrm{Na}^+/K^+$ pump consumes approximately $30$--$40\%$ of the total ATP produced by a resting neuron and $10$--$20\%$ of total body ATP at rest.

Worked Example: Resting Membrane Potential Calculation

Using the Goldman equation (constant field approximation) for a neuron:

$V_m = \frac{RT}{F}\ln\frac{P_{\mathrm{Na}}[\mathrm{Na}^+]_{out} + P_{\mathrm{K}}[\mathrm{K}^+]_{out} + P_{\mathrm{Cl}}[\mathrm{Cl}^-]_{in}}{P_{\mathrm{Na}}[\mathrm{Na}^+]_{in} + P_{\mathrm{K}}[\mathrm{K}^+]_{in} + P_{\mathrm{Cl}}[\mathrm{Cl}^-]_{out}}$

At $37^\circ\mathrm{C}$ ($RT/F = 26.7\;\mathrm{mV}$), with:

• $[\mathrm{Na}^+]_{out} = 145\;\mathrm{mM}$, $[\mathrm{Na}^+]_{in} = 12\;\mathrm{mM}$
• $[\mathrm{K}^+]_{out} = 4\;\mathrm{mM}$, $[\mathrm{K}^+]_{in} = 155\;\mathrm{mM}$
• $[\mathrm{Cl}^-]_{out} = 110\;\mathrm{mM}$, $[\mathrm{Cl}^-]_{in} = 4\;\mathrm{mM}$
• $P_{\mathrm{Na}} = 0.04$, $P_{\mathrm{K}} = 1.0$, $P_{\mathrm{Cl}} = 0.45$

Calculate the resting membrane potential.

Solution

Numerator $= 0.04 \times 145 + 1.0 \times 4 + 0.45 \times 4 = 5.8 + 4.0 + 1.8 = 11.6$

Denominator $= 0.04 \times 12 + 1.0 \times 155 + 0.45 \times 110 = 0.48 + 155 + 49.5 = 204.98$

$V_m = 26.7 \times \ln(11.6 / 204.98) = 26.7 \times \ln(0.0566) = 26.7 \times (-2.871) = -76.6\;\mathrm{mV}$

This is close to the typical neuronal resting potential of $-70\;\mathrm{mV}$. The difference reflects the contribution of the electrogenic $\mathrm{Na}^+/K^+$ pump (approximately $-3$ to $-5\;\mathrm{mV}$), which is not included in the Goldman equation.