Evolution in Depth

1. Evidence for Evolution

Fossil Evidence

The fossil record provides direct evidence of organisms from past geological eras, preserved in sedimentary rock.

Key observations:

Fossils in deeper (older) rock layers are progressively simpler and differ more from modern organisms than those in shallower (younger) layers.
Transitional fossils exhibit features intermediate between ancestral and descendant groups:

Transitional fossil	Significance
*Tiktaalik roseae* (375 Mya)	Intermediate between lobe-finned fish and early tetrapods: fish-like scales and fins with limb-like bones and a functional neck.
*Archaeopteryx lithographica* (150 Mya)	Dinosaur-like skeleton with teeth and long bony tail, but feathered wings and a furcula (wishbone). Links reptiles and birds.
*Australopithecus afarensis* ("Lucy", 3.2 Mya)	Bipedal pelvis and femur with ape-like cranial capacity ( $\approx 400\;\mathrm{cm}^3$ ). Ancestor of the genus Homo.
*Ambulocetus natans* (49 Mya)	"Walking whale": amphibious predator with functional limbs linking terrestrial artiodactyls and fully aquatic cetaceans.
*Tetrapod tracks* (395 Mya, Poland)	Fossilised footprints pre-dating the earliest tetrapod body fossils, showing animals walked on land earlier than skeletal evidence suggested.

Limitations of the fossil record:

Incompleteness: fossilisation requires rapid burial in anoxic, mineral-rich conditions (rare). Soft-bodied organisms (bacteria, worms, jellyfish) rarely fossilise. Most organisms that have ever lived left no fossil record.
Bias: marine organisms with hard parts (shells, bones) are overrepresented; tropical and forest environments have poor fossil preservation.
Resolution: speciation events (which may occur in small, isolated populations over thousands of years) may not be captured in the geological record.

Anatomical Evidence

Homologous structures: similar anatomy in different species due to common ancestry, despite potentially different functions.

Structure	Species	Common ancestry	Function in each species
Pentadactyl limb	Human, bat, whale, horse	Common tetrapod ancestor	Grasping, flying, swimming, running
Vertebral column	All vertebrates	Common chordate ancestor	Axial support and protection of spinal cord
Pharyngeal (gill) arches	Fish, amphibian, reptile, mammal (embryonic)	Common chordate ancestor	Gill support (fish); modified into jaws, ear ossicles, hyoid (mammals)

The pentadactyl limb is a particularly powerful example: the same basic bone arrangement (humerus, radius and ulna, carpals, metacarpals, phalanges) is adapted for vastly different functions --- grasping, flying, swimming, digging --- indicating modification from a common ancestral structure, not independent origin.

Analogous structures: similar function but different anatomical origin and embryological development. Result from convergent evolution (see Section 5).

Analogous structure	Species	Function	Different origins
Wings	Insect, bird, bat	Flight	Insect: exoskeletal extensions; bird/bat: modified vertebrate forelimb
Eye	Cephalopod (octopus), vertebrate	Image formation	Cephalopod: pinhole camera; vertebrate: lens eye; different embryological development
Streamlined body	Shark (fish), dolphin (mammal)	Swimming	Different skeletal structures; convergent on hydrodynamic form

Molecular Evidence

Comparing DNA, RNA, and protein sequences between species provides quantitative evidence for evolutionary relationships.

Cytochrome c comparisons: cytochrome c is a highly conserved protein involved in the electron transport chain. The amino acid sequence differs by a known number of residues between species:

Species compared to human	Amino acid differences (out of 104)
Chimpanzee	0
Rhesus monkey	1
Dog	10
Horse	12
Chicken	13
Tuna	21
Yeast	44

The pattern of differences is consistent with the accepted phylogeny: more closely related species share more similar sequences because less time has elapsed for mutations to accumulate.

Molecular clocks: the rate of molecular evolution (number of mutations per unit time) is approximately constant for neutral mutations. By counting sequence differences between two species and calibrating with the fossil record, the time since divergence can be estimated:

$t = \frac{d}{2r}$

where $d$ is the number of substitutions per site between the two sequences, $r$ is the mutation rate per site per year, and the factor of $2$ accounts for divergence along both lineages.

Limitations: mutation rates vary between genes (functional constraints), between lineages (generation time effects), and over time. Molecular clocks must be calibrated with multiple fossil dates and should be applied cautiously.

Embryological Evidence

Early embryonic stages of vertebrates share remarkable similarities:

Pharyngeal pouches: present in all vertebrate embryos (become gills in fish; modified into ear bones, tonsils, thymus in mammals).
Post-anal tail: present in all vertebrate embryos (regresses in humans, retained in other species).
Notochord: a flexible rod of cells that serves as the embryonic axial skeleton; replaced by vertebrae in most vertebrates but persists as the nucleus pulposus of intervertebral discs.

These shared embryonic features reflect common ancestry. "Ontogeny recapitulates phylogeny" (Haeckel's biogenetic law) is an oversimplification, but embryological similarities do provide evidence for shared evolutionary origins.

Biogeographical Evidence

The geographic distribution of species provides evidence for evolution:

Island biogeography: species on oceanic islands resemble those on the nearest mainland (e.g., Darwin's finches resemble South American finches; Galapagos iguanas resemble South American iguanas) but have evolved distinct adaptations.
Convergent evolution in similar habitats: the marsupial mammals of Australia (marsupial wolf, marsupial mole, marsupial flying phalanger) occupy ecological niches similar to placental mammals on other continents, demonstrating that similar selective pressures produce similar adaptations independently.
Fossil distribution: the distribution of Glossopteris fossils (a Permian fern) across South America, Africa, India, Antarctica, and Australia supports continental drift and explains why identical species are found on separated landmasses.

2. Natural Selection in Detail

Types of Natural Selection

Directional Selection

Favours one extreme phenotype, shifting the population mean in that direction.

$\Delta\bar{z} = \frac{G \cdot S}{\bar{w}}$

where $\Delta\bar{z}$ is the change in the mean phenotype, $G$ is the additive genetic variance, $S$ is the selection differential (difference between the mean of selected parents and the population mean), and $\bar{w}$ is the mean fitness.

Examples:

Antibiotic resistance: antibiotic treatment kills susceptible bacteria, selecting for resistant individuals. The population mean shifts toward resistance.
Peppered moth (Biston betularia): industrial pollution darkened tree bark, favouring the melanic (dark) form over the light form. After pollution control, the direction reversed.
Giraffe neck length: longer necks provide a feeding advantage, favouring individuals with longer necks (though this example is debated --- current evidence suggests sexual selection may also play a role).

Stabilising Selection

Favours the intermediate phenotype, reducing variation. Occurs when extreme phenotypes have lower fitness.

Examples:

Human birth weight: babies with very low or very high birth weights have higher mortality; intermediate weights ( $\approx 3.0$ -- $4.0\;\mathrm{kg}$ ) have the highest survival rate.
Sickle-cell allele in malaria-endemic regions: homozygous normal ( $HbA/HbA$ ) are susceptible to malaria; homozygous sickle-cell ( $HbS/HbS$ ) have sickle-cell disease; heterozygous ( $HbA/HbS$ ) have mild or asymptomatic malaria resistance (balancing selection --- a form of stabilising selection maintaining polymorphism).
Clutch size in birds: too few eggs = low reproductive output; too many eggs = insufficient food per chick, high chick mortality. Intermediate clutch sizes are favoured.

Disruptive Selection

Favours both extremes over the intermediate phenotype, potentially leading to a bimodal distribution and eventually speciation.

Examples:

African seedcracker finch (Pyrenestes ostrinus): two beak sizes (large and small) are favoured because the available seeds are either hard or soft; intermediate beak sizes are inefficient at cracking either type.
Darwin's finches on the Galapagos: during drought, birds with very large or very small beaks have an advantage (large beaks crack hard seeds; small beaks handle small soft seeds), while intermediate beaks are less efficient.

Sexual Selection

A form of natural selection where individuals with certain traits are more successful at obtaining mates.

Intrasexual selection: competition within one sex (usually males) for access to mates. Leads to weapons (antlers in deer, large body size) and aggressive behaviour.
Intersexual selection: one sex (usually females) chooses mates based on preferred traits. Leads to elaborate ornaments (peacock's tail, bird of paradise plumage, frog calls) that may reduce survival but increase reproductive success.

Handicap principle (Zahavi, 1975): elaborate ornaments are honest signals of fitness because only healthy, well-nourished individuals can afford the cost of producing and maintaining them.

Frequency-Dependent Selection

The fitness of a phenotype depends on its frequency in the population.

Negative frequency-dependent selection: rare phenotypes have a fitness advantage (e.g., side-blotched lizards: orange males are successful when rare, but lose advantage when common because blue males adapt to their strategy).
Positive frequency-dependent selection: common phenotypes have a fitness advantage (e.g., Mullerian mimicry in toxic species: predators learn to avoid the common warning pattern).

3. Speciation

Allopatric Speciation

The most common mechanism. Populations are separated by a geographic barrier; different selection pressures, genetic drift, and mutation in each population lead to reproductive isolation.

Stages:

Geographic isolation: a physical barrier (mountain range, river, ocean, glacier) splits the population.
Independent evolution: each population experiences different mutations, selection pressures, and genetic drift.
Reproductive isolation accumulates: over many generations, pre-zygotic and/or post-zygotic isolating mechanisms evolve.
Even if the populations come back into contact, they cannot or do not interbreed: they are now separate species.

Evidence: the Kaibab squirrel (Sciurus kaibabensis) and Abert's squirrel (Sciurus aberti) are separated by the Grand Canyon and have evolved distinct coat colours and behaviours despite their geographic proximity.

Sympatric Speciation

New species arise without geographic separation.

Mechanisms:

Polyploidy (most common in plants): an error in meiosis produces offspring with extra sets of chromosomes (e.g., tetraploid, $4n$ ). The polyploid individual can self-fertilise or mate with other polyploids, but cannot produce fertile offspring with the original diploid ( $2n$ ) population due to unequal chromosome pairing in meiosis. This creates instant reproductive isolation. Approximately $30$ -- $80\%$ of flowering plant species are polyploid.
Ecological speciation: different ecological niches within the same geographic area select for different traits. Example: Rhagoletis pomonella (apple maggot fly) originally infested hawthorn trees; after apple trees were introduced to North America ( $\approx 200$ years ago), a host race adapted to apples. The two populations now have different fruiting times (apple vs hawthorn) and show partial reproductive isolation based on host preference.
Sexual selection: different female mating preferences within a population can drive divergence.

Parapatric Speciation

Populations are adjacent with a narrow hybrid zone; gene flow occurs at the boundary but is reduced by selection against hybrids.

Example: Anthoxanthum odoratum (sweet vernal grass) grows on mine-contaminated soil (high copper tolerance) and uncontaminated soil. Flowering time differs between the two populations, reducing gene flow. Hybrid individuals have intermediate tolerance and reduced fitness in both habitats.

Reproductive Isolating Mechanisms

Pre-zygotic barriers (prevent mating or fertilisation):

Type	Example
Temporal	Different breeding seasons (e.g., spotted skunks: summer vs winter breeding populations).
Habitat	Different habitats within the same area (e.g., jungle vs open-field mosquito populations in Anopheles gambiae complex).
Behavioural	Different courtship rituals or songs (e.g., crickets, birds).
Mechanical	Incompatible reproductive structures (e.g., damselfly species with differently shaped claspers).
Gametic	Sperm cannot fertilise egg (e.g., sea urchin species with incompatible bindin proteins).

Post-zygotic barriers (reduce fitness of hybrid offspring):

Type	Example
Hybrid inviability	Hybrid zygotes fail to develop or have reduced survival.
Hybrid sterility	Hybrids are viable but sterile (e.g., mule: horse $\times$ donkey, $2n = 63$ ; uneven chromosome pairing in meiosis).
Hybrid breakdown	First-generation hybrids are fertile, but the F2 generation has reduced fitness.

4. Adaptive Radiation and Convergent Evolution

Adaptive Radiation

The rapid diversification of a single ancestral species into many morphologically and ecologically distinct species, often following the colonisation of a new habitat with diverse ecological niches.

Classic examples:

Darwin's finches on the Galapagos Islands: a single ancestral finch species from South America gave rise to $13$ -- $17$ species with diverse beak shapes and sizes adapted to different food sources (seeds, insects, cactus flowers, blood, tools). Peter and Rosemary Grant's long-term study (1973--present) documented natural selection on beak size in real time during drought years.
Hawaiian honeycreepers: a single coloniser gave rise to over $50$ species with diverse beak morphologies (insect-picking, nectar-sipping, seed-cracking, wood-boring), many of which are now extinct due to human activity.
Cichlid fish in East African Rift Lakes (Victoria, Malawi, Tanganyika): hundreds of species evolved within individual lakes in $< 15000$ years, occupying diverse trophic niches. Speciation was driven by sexual selection (female mate choice based on male colouration) and ecological specialisation.
Mammalian radiation after the Cretaceous-Paleogene extinction: the extinction of non-avian dinosaurs $66$ Mya vacated many ecological niches, allowing mammals to rapidly diversify into the orders we see today (primates, carnivores, cetaceans, rodents, etc.).

Conditions favouring adaptive radiation:

A new habitat or ecological opportunity with unoccupied niches.
Evolutionary innovations (key adaptations that open new niches, e.g., amniotic egg, flight, photosynthesis).
Mass extinction events that remove competitors and create vacant niches.

Convergent Evolution

Independent evolution of similar traits in distantly related species in response to similar environmental pressures.

Examples:

Trait	Species pairs/groups	Selective pressure
Streamlined body	Shark (fish), dolphin (mammal), ichthyosaur (reptile)	Drag reduction in aquatic locomotion
Camera eye	Cephalopod (octopus), vertebrate (human)	Image formation; detection of light/predators
Wings	Bird, bat, pterosaur, insect	Flight
C4 photosynthesis	Maize (grass), sorghum (grass), amaranth (dicot)	High temperature, high light, low $\mathrm{CO}_2$ efficiency
Echolocation	Bats (microchiroptera), toothed whales	Navigation and prey detection in darkness
Antifreeze proteins	Arctic cod, Antarctic notothenioid fish	Subzero water temperatures

Distinguishing homology from convergence: homologous structures share the same underlying anatomical plan and embryological origin despite different functions; analogous structures have different anatomical origins but similar functions. Molecular evidence can distinguish them: homologous traits often share similar developmental genes (e.g., Pax6 for eye development in both vertebrates and cephalopods); convergent traits arise from different genetic mechanisms.

5. Gradualism vs Punctuated Equilibrium

Phyletic Gradualism (Darwinian Gradualism)

Evolution occurs gradually and continuously over long periods.
Large morphological changes result from the accumulation of many small changes over geological time.
The fossil record should show smooth, continuous transitional forms.

Support: some fossil sequences show gradual change (e.g., foraminifera, trilobite ontogeny).

Punctuated Equilibrium (Gould and Eldredge, 1972)

Species experience long periods of stasis (little morphological change) punctuated by brief periods of rapid speciation.
Most morphological change occurs during the speciation event (typically in small, isolated populations over $10^4$ -- $10^5$ years).
The fossil record appears "jerky" because the transitional forms are rare and geographically localised.

Support: the fossil record often shows species appearing suddenly (in geological terms) and persisting unchanged for millions of years. For example, Trilobites in the Burgess Shale show well-defined species with little gradual transition between them.

Current synthesis: both patterns occur. Gradual change is well-documented in some lineages; punctuated equilibrium explains the apparent gaps in the fossil record. The two models are not mutually exclusive.

6. Hardy-Weinberg Equilibrium (Extended)

The Principle

For a population with two alleles ( $A$ and $a$ ) at a single locus, with frequencies $p$ and $q$ :

$p + q = 1$ $p^2 + 2pq + q^2 = 1$

Conditions for equilibrium:

No mutations (mutation rate $\mu \approx 0$ ).
No migration (gene flow, $m = 0$ ).
Random mating (no assortative or disassortative mating).
Infinite population size (no genetic drift).
No natural selection (all genotypes have equal fitness).

Testing Hardy-Weinberg with Multiple Alleles

For a locus with $n$ alleles with frequencies $p_1, p_2, \ldots, p_n$ :

$\sum_{i=1}^{n} p_i = 1$

The expected frequency of homozygote $p_i^2$ and heterozygote $2p_ip_j$ for all pairs.

Example: ABO blood group with three alleles ( $I^A$ , $I^B$ , $i$ ): $p_{I^A} + p_{I^B} + p_i = 1$ $f(I^AI^A) = p_{I^A}^2, \quad f(I^AI^B) = 2p_{I^A}p_{I^B}, \quad f(I^Ai) = 2p_{I^A}p_i, \quad \ldots$

Effects of Evolutionary Forces on Hardy-Weinberg

Mutation

Mutation introduces new alleles and changes allele frequencies, but at a very slow rate ( $\mu \approx 10^{-5}$ to $10^{-6}$ per generation per locus). Mutation alone is too slow to drive significant evolutionary change, but it provides the raw genetic variation upon which selection and drift act.

Migration (Gene Flow)

The introduction of alleles from another population changes allele frequencies:

$\Delta p = m(p_m - p_r)$

where $m$ is the migration rate, $p_m$ is the allele frequency in the migrant population, and $p_r$ is the allele frequency in the resident population. Gene flow tends to homogenise allele frequencies between populations and counteracts the effects of selection and drift.

Genetic Drift

Random fluctuations in allele frequencies due to sampling error in finite populations.

Strongest in small populations: in a population of $N$ diploid individuals, the variance in allele frequency change per generation is $\sigma^2 = \frac{p(1-p)}{2N}$ .
Bottleneck effect: a sharp reduction in population size (e.g., due to natural disaster, hunting) reduces genetic diversity. The surviving population has a non-representative allele frequency distribution. Example: northern elephant seals were hunted to $\approx 20$ individuals in the 1890s; modern populations have very low genetic diversity compared with southern elephant seals.
Founder effect: a small group establishes a new population, carrying only a subset of the source population's genetic diversity. Example: the Amish population of Pennsylvania has a high frequency of Ellis-van Creveld syndrome (a recessive dwarfism) due to the small number of founders.

Non-Random Mating

Assortative mating: individuals preferentially mate with similar (positive assortative) or dissimilar (negative assortative) partners. Positive assortative mating increases homozygosity without changing allele frequencies.
Inbreeding: mating between closely related individuals increases homozygosity and exposes recessive deleterious alleles. The inbreeding coefficient ( $F$ ) measures the probability that two alleles at a locus in an individual are identical by descent.

Natural Selection

Selection changes allele frequencies by differential reproductive success:

Selection type	Effect on allele frequencies	Effect on population variance
Directional	Favourable allele increases toward fixation	Initially decreases
Stabilising	Intermediate phenotype favoured; extreme alleles removed	Decreases
Disruptive	Extreme phenotypes favoured; intermediate alleles removed	Increases
Balancing (overdominance)	Heterozygote has highest fitness (e.g., sickle-cell trait)	Maintains; polymorphism
Frequency-dependent	Fitness depends on allele frequency; maintains oscillation	Variable

Fitness and Selection Coefficients

Fitness ( $w$ ): the relative reproductive success of a genotype, normalised so that the fittest genotype has $w = 1.0$ .

Selection coefficient ( $s$ ): the reduction in fitness of a genotype relative to the fittest: $w = 1 - s$ .

Example: if the fitness of genotypes $AA$ , $Aa$ , and $aa$ are $1.0$ , $1.0$ , and $0.8$ : $s = 0.2$ for the $aa$ genotype. The selection acts against the $aa$ homozygote.

Rate of change of a deleterious allele under selection: $\Delta q \approx -\frac{spq^2}{1 - sq^2}$

For a completely recessive deleterious allele ( $s$ against $aa$ ), the allele is sheltered in heterozygotes and is eliminated very slowly when rare ( $\Delta q \propto q^2$ ).

7. Phylogenetic Trees

Construction

Phylogenetic trees represent evolutionary relationships among species based on shared derived characters (synapomorphies).

Cladistics (Hennig, 1966): groups organisms into clades (monophyletic groups) based on shared derived characteristics. A clade includes an ancestor and all of its descendants.

Steps in constructing a phylogenetic tree:

Select characters (morphological, molecular, or both).
Code characters as ancestral (plesiomorphic) or derived (apomorphic).
Construct a character matrix (taxa $\times$ characters).
Identify shared derived characters that define clades.
Use parsimony (prefer the tree requiring the fewest evolutionary changes) or maximum likelihood / Bayesian methods (find the tree most likely given the data and an explicit model of evolution).

Types of Trees

Rooted tree: includes a common ancestor and shows the direction of evolution (usually rooted using an outgroup --- a taxon known to be outside the group of interest).
Unrooted tree: shows relationships but not evolutionary direction.
Ultrametric tree: branch lengths are proportional to time (requires molecular clock calibration).

Interpreting Phylogenetic Trees

Branching points (nodes) represent speciation events or common ancestors.
Tips (terminal nodes) represent extant or extinct species.
Branch lengths may represent the number of character changes (non-ultrametric) or time (ultrametric).
Closely related species share a more recent common ancestor (shorter branch to the most recent node they share).

Molecular Phylogenetics

DNA and protein sequences provide large datasets for phylogenetic analysis:

Sequence alignment: homologous sequences are aligned to identify conserved and variable positions (e.g., using ClustalW, MUSCLE).
Substitution models: mathematical models describe the probability of one nucleotide substituting for another (e.g., Jukes-Cantor, Kimura 2-parameter, GTR).
Tree-building methods:
- Maximum parsimony: minimises the total number of changes.
- Maximum likelihood: finds the tree that maximises the probability of the observed data given the substitution model.
- Bayesian inference: uses MCMC sampling to estimate the posterior probability distribution of trees.

8. Antibiotic Resistance as Evolution

Antibiotic resistance provides a clear, observable example of evolution by natural selection:

Variation exists: within a bacterial population, some individuals carry resistance genes (on plasmids, transposons, or chromosomes) due to prior mutations or horizontal gene transfer.
Selective pressure: antibiotic treatment kills susceptible bacteria ( $\approx 99.9\%$ ), creating intense selection for resistant individuals.
Differential survival: resistant bacteria survive and reproduce.
Allele frequency change: the resistance allele frequency increases from $< 0.1\%$ to $\approx 100\%$ in the population within a few generations.
Spread: resistance genes can spread horizontally between bacterial species via conjugation (plasmid transfer), transformation (uptake of free DNA), and transduction (bacteriophage transfer).

Mechanisms of resistance:

Mechanism	Example
Enzymatic inactivation	$\beta$ -lactamase hydrolyses penicillin; aminoglycoside-modifying enzymes.
Target modification	MRSA: altered penicillin-binding protein (PBP2a) with low affinity for $\beta$ -lactams.
Efflux pumps	Active transport of antibiotics out of the cell (e.g., tetracycline efflux).
Reduced permeability	Mutations reducing porin channel size (e.g., resistance to carbapenems in Pseudomonas).
Target bypass	Acquisition of a resistant enzyme variant (e.g., mecA gene for methicillin resistance).

Preventing resistance:

Complete the full antibiotic course (do not stop early).
Use antibiotics only when necessary (do not use for viral infections).
Rotate antibiotic classes in agriculture.
Develop new antibiotics and alternative therapies (phage therapy, antimicrobial peptides).

Common Pitfalls

Confusing homologous and analogous structures: homologous = shared ancestry, different function (pentadactyl limb); analogous = different ancestry, similar function (bird wing vs insect wing).
Stating that "evolution is just a theory": in science, "theory" means a well-substantiated explanation supported by extensive evidence, not a guess.
Describing natural selection as "survival of the fittest" without defining fitness: fitness is measured by reproductive success, not strength or size.
Assuming that "individuals evolve": evolution occurs at the population level through changes in allele frequency. Individuals do not evolve; populations do.
Misapplying Hardy-Weinberg: real populations never meet all five conditions. The H-W principle is a null model used to detect evolutionary forces, not a description of nature.
Confusing allopatric and sympatric speciation: allopatric requires geographic separation; sympatric occurs without it (polyploidy, ecological speciation).
Assuming convergent evolution produces identical structures: the wings of birds, bats, and insects are analogous (same function) but have fundamentally different anatomical origins.

Practice Problems

Question 1: Hardy-Weinberg with Selection

In a population of $10000$ plants, flower colour is controlled by a single gene with two alleles: $R$ (red, dominant) and $r$ (white, recessive). The current genotype frequencies are: $f(RR) = 0.49$ , $f(Rr) = 0.42$ , $f(rr) = 0.09$ . A disease selectively kills white-flowered plants ( $rr$ ) with $50\%$ mortality ( $w_{rr} = 0.5$ ), while red-flowered plants are unaffected ( $w_{RR} = w_{Rr} = 1.0$ ). Calculate the allele frequencies after one generation of selection.

Answer

Initial allele frequencies: $p = 0.49 + 0.5 \times 0.42 = 0.49 + 0.21 = 0.70$ $q = 0.09 + 0.5 \times 0.42 = 0.09 + 0.21 = 0.30$

Mean fitness: $\bar{w} = p^2 w_{RR} + 2pq \cdot w_{Rr} + q^2 w_{rr} = (0.49)(1.0) + (0.42)(1.0) + (0.09)(0.5) = 0.49 + 0.42 + 0.045 = 0.955$

Genotype frequencies after selection: $f'(RR) = \frac{0.49}{0.955} = 0.5131$ $f'(Rr) = \frac{0.42}{0.955} = 0.4398$ $f'(rr) = \frac{0.045}{0.955} = 0.0471$

New allele frequencies: $p' = 0.5131 + 0.5 \times 0.4398 = 0.5131 + 0.2199 = 0.7330$ $q' = 0.0471 + 0.5 \times 0.4398 = 0.0471 + 0.2199 = 0.2670$

The $R$ allele frequency increased from $0.70$ to $0.733$ in one generation due to selection against the $rr$ genotype. The $r$ allele declines slowly because it is sheltered in heterozygotes.

Question 2: Molecular Clock Calculation

The cytochrome c protein has $104$ amino acids. Between humans and yeast, there are $44$ differences. If the average substitution rate for cytochrome c is $1$ amino acid change per $20$ million years per lineage, estimate the time since humans and yeast last shared a common ancestor.

Answer

Total substitutions = $44$ amino acid differences. Substitution rate = $1$ change per $20$ My per lineage.

Time since divergence: $t = \frac{d}{2r} = \frac{44}{2 \times (1/20)} = \frac{44}{0.1} = 440$ million years.

The estimated divergence time is $\approx 440$ Mya. This is consistent with the fossil and phylogenetic evidence that animals and fungi diverged approximately $400$ -- $600$ Mya. The approximation is reasonable given the simplifying assumptions (constant rate, no back-mutations, no saturation effects).

Question 3: Speciation and Reproductive Isolation

Two populations of a freshwater fish are separated by a waterfall. The upstream population lives in a cold, fast-flowing stream; the downstream population lives in a warm, slow-moving pond. After $50000$ years, the waterfall dries up and the two populations re-establish contact. However, they do not interbreed. Identify three possible reproductive isolating mechanisms that may have evolved, classifying each as pre-zygotic or post-zygotic. Explain how each could have arisen in this scenario.

Answer

Temporal isolation (pre-zygotic): the two habitats have different temperatures and seasonal patterns. The cold upstream population may breed in spring (triggered by temperature increase); the warm downstream population may breed year-round or at a different time. Even in contact, their breeding seasons do not overlap, preventing mating.
Behavioural isolation (pre-zygotic): in the fast-flowing stream, visual courtship displays may be ineffective; the upstream population may have evolved different mating calls or pheromones adapted to their environment. Females may no longer recognise the courtship signals of males from the other population.
Hybrid inviability or sterility (post-zygotic): if occasional interbreeding occurred before complete pre-zygotic isolation evolved, hybrid offspring may have had reduced fitness in either parental habitat (intermediate temperature tolerance, neither fully adapted to fast water nor still water). Over time, genetic incompatibilities (Dobzhansky-Muller incompatibilities) could have accumulated, making hybrids sterile or inviable.

Question 4: Antibiotic Resistance and Selection

A population of $10^8$ bacteria is treated with an antibiotic. The antibiotic kills $99.999\%$ of the bacteria. The resistant survivors carry a plasmid-borne resistance gene. After $8$ generations of binary fission, the population recovers to its original size. Calculate the number of survivors after treatment, the fraction carrying the resistance gene after recovery, and the minimum number of generations required for $99.9\%$ of the population to carry the resistance gene.

Answer

Survivors after treatment: $10^8 \times (1 - 0.99999) = 10^8 \times 10^{-5} = 1000$ bacteria. All $1000$ survivors carry the resistance gene.

After 8 generations of binary fission: Population size: $1000 \times 2^8 = 1000 \times 256 = 256000$ bacteria. All are descendants of the $1000$ resistant survivors, so $100\%$ carry the resistance gene. The population has not yet recovered to its original size of $10^8$ .

Generations for $99.9\%$ resistance in a population of $10^8$ : After $n$ generations, the resistant population = $1000 \times 2^n$ . We need $1000 \times 2^n = 0.999 \times 10^8 = 9.99 \times 10^7$ . $2^n = 99900$ $n = \frac{\ln(99900)}{\ln(2)} = \frac{11.513}{0.693} \approx 16.6$ , so $17$ generations.

After $17$ generations, $\approx 1000 \times 2^{17} = 1.31 \times 10^8$ bacteria, virtually all carrying the resistance gene. This demonstrates how quickly resistance can spread under antibiotic selection.

Question 5: Phylogenetic Tree Interpretation

A molecular phylogenetic tree of four species (A, B, C, D) based on a conserved gene shows the following topology: ((A, B), C), D; where D is the outgroup. The branch lengths (number of substitutions per site) are: A--B common ancestor = $0.02$ , A--B split to present = $0.01$ each, A--B/C common ancestor to C = $0.05$ , A--B/C/D common ancestor to A--B/C common ancestor = $0.03$ . (a) Which two species are most closely related? (b) Calculate the total evolutionary distance between A and C. (c) Explain why D is used as an outgroup.

Answer

(a) Species A and B are most closely related: they share the most recent common ancestor (the node separating them from C and D is the shallowest).

(b) Total distance A to C = path from A to A/B node ( $0.01$ ) + A/B node to A/B/C node ( $0.02$ ) + A/B/C node to C ( $0.05$ ) = $0.01 + 0.02 + 0.05 = 0.08$ substitutions per site.

(c) D is used as an outgroup because it is known (from other evidence) to be less closely related to A, B, and C than they are to each other. The outgroup "roots" the tree, allowing determination of the direction of character changes (ancestral vs derived) and providing a reference point for polarising the tree.

Worked Examples

Worked Example: Genetic Drift and the Bottleneck Effect

A population of $1000$ butterflies has two alleles at a wing-colour locus: $C$ (common, $p = 0.8$ ) and $c$ (rare, $q = 0.2$ ). A wildfire kills $95\%$ of the population, leaving $50$ survivors. Calculate the probability that the $c$ allele is completely lost in this bottleneck event.

Solution

The probability that allele $c$ is lost = probability that none of the $50$ survivors carries the $c$ allele.

Each individual has two alleles. The probability that a randomly chosen individual does NOT carry $c$ is the probability of being $CC$ : $p^2 = 0.8^2 = 0.64$ .

The probability that all $50$ survivors are $CC$ : $(0.64)^{50} = 1.27 \times 10^{-9}$

This is extremely unlikely. However, the probability that the $c$ allele frequency decreases significantly is much higher. The variance in allele frequency after a bottleneck is: $\sigma^2_q = \frac{pq}{2N_e} = \frac{0.8 \times 0.2}{2 \times 50} = \frac{0.16}{100} = 0.0016$

The standard deviation: $\sigma_q = \sqrt{0.0016} = 0.04$

After the bottleneck, the expected $c$ frequency is still $q = 0.2$ , but there is a $95\%$ chance it falls within $0.2 \pm 2(0.04) = 0.12$ to $0.28$ . In some bottleneck events, the $c$ allele frequency could change substantially, and over many repeated bottlenecks, the probability of eventual loss approaches $1$ .

Worked Example: Selection Against a Recessive Lethal Allele

Cystic fibrosis (CF) is caused by a recessive lethal allele ( $c$ ). The disease incidence in a population is $1$ in $2500$ ( $q^2 = 1/2500$ ). If modern medicine allows CF patients to survive to reproductive age (changing fitness from $w_{cc} = 0$ to $w_{cc} = 0.5$ ), predict the new equilibrium frequency of the $c$ allele after many generations.

Solution

Current equilibrium ( $w_{cc} = 0$ , lethal): $q^2 = 1/2500$ , so $q = 1/50 = 0.02$ . $p = 0.98$ . Carrier frequency: $2pq = 2 \times 0.98 \times 0.02 = 0.0392 \approx 1/25.5$ .

New equilibrium ( $w_{CC} = 1.0$ , $w_{Cc} = 1.0$ , $w_{cc} = 0.5$ ): At equilibrium, the frequency of the $c$ allele is maintained by the balance between mutation (introducing $c$ ) and selection (removing $c$ ). With $w_{cc} = 0.5$ :

$\hat{q} = \frac{\mu}{s}$ (approximation for a rare recessive deleterious allele, where $s$ is the selection coefficient against homozygotes)

With $s = 0.5$ : if $\mu \approx 10^{-6}$ (typical mutation rate), $\hat`\{q}` = 10^{-6} / 0.5 = 2 \times 10^{-6}$ , which is negligible. However, if the selection pressure is relaxed completely ( $s = 0$ , no fitness disadvantage), the allele frequency will remain at its current value ( $q = 0.02$ ) unless mutation rates are negligible.

With $w_{cc} = 0.5$ (partial selection), the equilibrium frequency will be between the two extremes. The frequency will increase from $0.02$ because homozygotes can now survive and reproduce. Using the equilibrium condition for a deleterious recessive:

$\hat{q} \approx \sqrt{\frac{\mu}{s}} = \sqrt{\frac{10^{-6}}{0.5}} = \sqrt{2 \times 10^{-6}} \approx 0.0014$

Wait --- this gives a lower frequency, which is incorrect. The issue is that $\mu/s$ applies when mutation is introducing the allele and selection is removing it. If we are starting from $q = 0.02$ and reducing selection from $s = 1$ to $s = 0.5$ , the frequency will actually increase because fewer $cc$ individuals are being removed.

The correct approach: with reduced selection, the equilibrium moves to a higher $q$ . The new equilibrium is approximately $q \approx \sqrt{\mu/s}$ , which with a smaller $s$ gives a larger $q$ . But this formula assumes mutation-selection balance. In reality, starting from $q = 0.02$ with relaxed selection, the allele frequency will increase over generations as $cc$ individuals survive and reproduce, approaching a new equilibrium that depends on the mutation rate.

Common Pitfalls (Expanded)

Confusing homologous and analogous structures: homologous = common ancestry (pentadactyl limb); analogous = similar function from convergent evolution (bird wing vs insect wing).
Stating that "evolution is just a theory": the term "theory" in science means a comprehensive explanation supported by extensive evidence.
Describing natural selection as "survival of the fittest": fitness = reproductive success, not physical strength. An organism that reproduces successfully is "fitter" than a stronger one that does not reproduce.
Assuming individuals evolve: evolution is a population-level process. Individuals do not evolve; populations change in allele frequency over generations.
Misapplying Hardy-Weinberg: real populations never meet all five conditions. H-W is a null model for detecting evolutionary forces.
Confusing allopatric and sympatric speciation: allopatric requires geographic separation; sympatric occurs without it.
Assuming molecular clocks are perfectly constant: mutation rates vary between lineages, genes, and over time. Calibration with multiple fossils is essential.

Exam-Style Problems

Problem 1: Extended Response -- Evidence for Evolution

Evaluate the relative strengths and limitations of four types of evidence for evolution: (a) fossil record, (b) comparative anatomy, (c) molecular biology, and (d) biogeography. For each type, provide a specific named example and discuss one limitation. Explain how multiple lines of evidence together provide a more robust argument for evolution than any single type alone.

Problem 2: Data Analysis -- Hardy-Weinberg and Selection

In a population of $2000$ snails, shell colour is controlled by two alleles: $B$ (brown, dominant) and $b$ (yellow, recessive). The observed numbers are: $B B = 720$ , $Bb = 960$ , $bb = 320$ . (a) Calculate the allele frequencies. (b) Test whether the population is in Hardy-Weinberg equilibrium using the chi-squared test ( $p = 0.05$ , critical value for $1$ df $= 3.84$ ). (c) If thrushes selectively prey on yellow snails ( $bb$ ) with $40\%$ mortality while brown snails are unaffected, calculate the new allele frequencies after one generation of selection.

Problem 3: Extended Response -- Speciation on Islands

A volcanic island emerges from the ocean $500000$ years ago and is colonised by a single species of beetle from the nearest mainland ( $200\;\mathrm{km}$ away). Today, the island has $12$ species of beetles, all descended from the original coloniser. (a) Explain the evolutionary processes that could have produced this diversity from a single ancestral species. (b) Discuss the role of adaptive radiation in this scenario. (c) Explain how reproductive isolating mechanisms could have evolved between the island species despite their geographic proximity. (d) Predict what would happen to these species if a land bridge connected the island to the mainland.

Problem 4: Extended Response -- Antibiotic Resistance as Evolution

Methicillin-resistant Staphylococcus aureus (MRSA) is a major hospital-acquired infection. (a) Describe the genetic and biochemical mechanisms by which MRSA resists methicillin and related $\beta$ -lactam antibiotics. (b) Explain how the overuse of antibiotics in hospitals and agriculture has contributed to the spread of resistance. (c) Evaluate two strategies for controlling the spread of antibiotic resistance, discussing the evolutionary principles behind each.

Problem 5: Quantitative -- Molecular Clock and Divergence

Two species of fruit fly (Drosophila simulans and D. melanogaster) differ by $60$ nucleotide substitutions in a $1000\;\mathrm{bp}$ region of the Adh gene. The substitution rate for this gene is estimated at $1.5 \times 10^{-8}$ substitutions per site per year. (a) Calculate the time since divergence. (b) Explain two assumptions of the molecular clock that may not hold for these species. (c) If a third species (D. yakuba) differs from D. melanogaster by $90$ substitutions in the same region, construct the most parsimonious phylogenetic tree and identify the outgroup.

If You Get These Wrong, Revise:

Genetics and inheritance --> Review ./genetics
Molecular biology and DNA --> Review ./molecular-biology
Ecology and populations --> Review ./ecology
Classification and taxonomy --> Review ./ecology
Genetic engineering techniques --> Review ./genetics-advanced

Additional Worked Examples

Worked Example: Hardy-Weinberg with Selection

In a population of moths, the allele for dark colour ( $D$ ) is dominant over light colour ( $d$ ). The initial allele frequencies are $p = 0.2$ and $q = 0.8$ . Light-coloured moths have a survival rate of $60\%$ (selection coefficient $s = 0.4$ against the recessive phenotype). (a) Calculate the genotype frequencies before selection. (b) Calculate the relative fitness of each genotype. (c) Calculate the allele frequencies after one generation of selection. (d) How many generations would it take for the frequency of $d$ to drop below $0.5$ ?

Solution

(a) Before selection: $p = 0.2$ , $q = 0.8$ . $P(DD) = p^2 = 0.04$ $P(Dd) = 2pq = 0.32$ $P(dd) = q^2 = 0.64$

(b) Relative fitness ( $w$ ):

$DD$ : $w = 1.0$ (dark moths have no disadvantage)
$Dd$ : $w = 1.0$ (dominant phenotype, no disadvantage)
$dd$ : $w = 1 - s = 1 - 0.4 = 0.6$

(c) Mean fitness: $\bar{w} = p^2 w_{DD} + 2pq w_{Dd} + q^2 w_{dd}$ $= 0.04(1.0) + 0.32(1.0) + 0.64(0.6) = 0.04 + 0.32 + 0.384 = 0.744$

After selection, the frequency of $d$ is: $q' = \frac{q^2 w_{dd} + pq w_{Dd}}{\bar{w}} = \frac{0.64(0.6) + 0.2 \times 0.8(1.0)}{0.744}$ $= \frac{0.384 + 0.16}{0.744} = \frac{0.544}{0.744} = 0.731$

$p' = 1 - q' = 0.269$

After one generation: $p = 0.269$ , $q = 0.731$ .

(d) We need to track $q$ over successive generations. Using the recurrence relation: $q' = \frac{q^2(1-s) + pq}{\bar{w}}$ where $\bar{w} = 1 - sq^2$ .

For selection against a recessive, the change in $q$ per generation is approximately: $\Delta q \approx -\frac{spq^2}{1 - sq^2}$

This is slow when $q$ is small (because most $d$ alleles are hidden in heterozygotes). Let me compute successively:

Gen 0: $q = 0.8$ , $p = 0.2$ , $\bar{w} = 1 - 0.4(0.64) = 0.744$ $q' = (0.64 \times 0.6 + 0.16)/0.744 = 0.544/0.744 = 0.731$

Gen 1: $q = 0.731$ , $p = 0.269$ , $\bar{w} = 1 - 0.4(0.534) = 0.786$ $q' = (0.534 \times 0.6 + 0.269 \times 0.731)/0.786 = (0.321 + 0.197)/0.786 = 0.518/0.786 = 0.659$

Gen 2: $q = 0.659$ , $\bar{w} = 1 - 0.4(0.434) = 0.826$ $q' = (0.434 \times 0.6 + 0.341 \times 0.659)/0.826 = (0.261 + 0.225)/0.826 = 0.486/0.826 = 0.588$

Gen 3: $q = 0.588$ , $\bar{w} = 1 - 0.4(0.346) = 0.862$ $q' = (0.346 \times 0.6 + 0.412 \times 0.588)/0.862 = (0.208 + 0.242)/0.862 = 0.450/0.862 = 0.522$

So after approximately $4$ generations, $q < 0.5$ .

Worked Example: Genetic Drift in a Small Population

A small island population of $20$ birds (10 males, 10 females) has an allele $A$ at frequency $p = 0.7$ . (a) Calculate the expected heterozygosity ( $H_e = 2pq$ ). (b) Calculate the effective population size ( $N_e$ ). (c) Calculate the expected heterozygosity after 5 generations of drift. (d) Explain why genetic drift has a stronger effect in small populations.

Solution

(a) $H_e = 2pq = 2 \times 0.7 \times 0.3 = 0.42$ .

(b) For a population with equal sex ratio: $N_e = \frac{4N_m N_f}{N_m + N_f} = \frac{4 \times 10 \times 10}{10 + 10} = \frac{400}{20} = 20$ . (Since $N_m = N_f$ , $N_e = N = 20$ .)

(c) Heterozygosity declines each generation by a factor of $(1 - \frac{1}{2N_e})$ : $H_t = H_0 \times (1 - \frac{1}{2N_e})^t = 0.42 \times (1 - \frac{1}{40})^5 = 0.42 \times (0.975)^5$ $(0.975)^5 = 0.881$ $H_5 = 0.42 \times 0.881 = 0.370$

After 5 generations, heterozygosity has declined from $0.42$ to $0.370$ (a loss of approximately $12\%$ ).

(d) Genetic drift is the random fluctuation of allele frequencies due to sampling error in each generation. In small populations, the sampling error is proportionally larger (each individual represents a larger fraction of the gene pool), so allele frequencies fluctuate more dramatically. In large populations, sampling effects average out. Genetic drift can lead to fixation or loss of alleles regardless of their adaptive value, reducing genetic variation. This is why small, isolated populations are more vulnerable to inbreeding depression and have reduced evolutionary potential.

Worked Example: Constructing a Phylogenetic Tree

Five species share the following presence/absence of derived characters (1 = present, 0 = absent):

Character	Species A	Species B	Species C	Species D
1	1	1	1	1
2	1	1	0	0
3	1	1	1	0
4	1	0	1	0
5	0	1	0	1

(a) Construct the most parsimonious cladogram. (b) Identify all synapomorphies (shared derived characters) at each node. (c) Identify any homoplasies (convergent evolution or reversals).

Solution

(a) Working from the outgroup (Species E, all characters = 0):

Character 1 is shared by A, B, C, D -- this is the first split (E branches off first).

Remaining group: A, B, C, D all have character 1.

Character 2: present in A and B only. This groups A + B together. Character 5: present in B and D only. This is a homoplasy (see below).

Character 3: present in A, B, C (not D). This groups A, B, C together. Character 4: present in A and C (not B). This is a homoplasy.

Most parsimonious tree (based on fewest changes):

         ┌── A (chars 2, 3, 4)
    ┌────┤
    │    └── B (chars 2, 3, 5)
────┤
    │    ┌── C (char 3, 4)
    └────┤
         └── D (char 5)

E (outgroup)

Node 1 (root): splits E from \\{A,B,C,D\\}. Synapomorphy: character 1. Node 2: splits \\{A,B\\} from \\{C,D\\}. Synapomorphy: character 2. Node 3: splits A from B. No unambiguous synapomorphy (character 4 in A and character 5 in B are autapomorphies or homoplasies). Node 4: splits C from D. No clear synapomorphy at this node.

(b) Synapomorphies:

Node 1 (A+B+C,D vs E): character 1
Node 2 (A+B vs C+D): character 2

Character 3 appears in A, B, C but not D. If the tree groups (A,B) and (C,D), then character 3 either evolved independently in the A+B clade and in C (convergent evolution), or was present in the ancestor of A+B+C+D and lost in D (reversal).
Character 4 appears in A and C but not B. This requires either convergent evolution or a reversal.
Character 5 appears in B and D but not A or C. This requires convergent evolution (independent gains) since B and D are on different branches.

The most parsimonious tree minimises the number of homoplasies. The tree above requires 3 homoplasies (characters 3, 4, 5). Alternative trees would need to be evaluated to find the one requiring the fewest total character state changes.

Worked Example: Speciation by Polyploidy

A diploid plant species has $2n = 14$ ( $n = 7$ ). (a) An individual undergoes autopolyploidy. What is the chromosome number of the tetraploid offspring? (b) Explain why the tetraploid is reproductively isolated from the diploid parent. (c) If the diploid parent produces gametes with $n = 7$ and the tetraploid produces gametes with $n = 14$ , what chromosome number would a triploid hybrid have, and why would it be sterile?

Solution

(a) Tetraploid: $4n = 28$ . The individual has 4 copies of each of the 7 chromosomes.

(b) The tetraploid is reproductively isolated because:

Diploid gametes: $n = 7$
Tetraploid gametes: $n = 14$
Hybrid offspring: $7 + 14 = 21$ (triploid)
During meiosis in the triploid, chromosomes cannot pair properly (there are 3 copies of each chromosome instead of 2). This leads to unbalanced gametes, resulting in sterility.
This is post-zygotic reproductive isolation (hybrid sterility).

(c) Triploid hybrid: $3n = 21$ . During meiosis, the 3 homologous chromosomes attempt to pair, but synapsis is irregular. Some chromosomes may form trivalents (3 chromosomes paired), others may form bivalent + univalent (2 paired + 1 unpaired). This leads to gametes with random numbers of chromosomes (7, 8, 9, etc.), most of which are non-viable. The triploid is therefore sterile.

This mechanism (polyploidy leading to instant speciation) is common in plants and rare in animals. Approximately $70\%$ of flowering plant species have undergone at least one polyploidisation event in their evolutionary history.

Worked Example: Bottleneck Effect and Genetic Diversity

A population of $10\,000$ individuals is reduced to $50$ individuals by a natural disaster (population bottleneck). Before the bottleneck, the frequency of a neutral allele $B$ is $0.3$ . (a) What is the probability that allele $B$ is lost from the population during the bottleneck? (b) After the bottleneck, the population recovers to $10\,000$ individuals. Explain why the genetic diversity of the recovered population is lower than the original. (c) The bottleneck population has a higher frequency of a rare genetic disease (autosomal recessive, allele $d$ ). Explain this observation.

Solution

(a) The probability that allele $B$ is lost depends on the sampling process. If we model this as a binomial sampling of $2N_e = 100$ gene copies (from the $50$ surviving individuals), the probability that zero copies of $B$ are sampled is $(1 - 0.3)^{100} = 0.7^{100} \approx 3.2 \times 10^{-16}$ .

However, this calculation assumes the bottleneck survivors are a perfectly random sample, which is unlikely. In practice, the founder effect during bottlenecks can dramatically change allele frequencies or eliminate alleles. The key point is that rare alleles (low frequency) are much more likely to be lost than common alleles. If $B$ had frequency $0.01$ , $P(\text{loss}) = 0.99^{100} \approx 0.366$ ( $37\%$ ).

(b) The bottleneck reduces the number of alleles in the population because many alleles present in the original $10\,000$ individuals are not represented among the $50$ survivors. When the population recovers, it can only regenerate the alleles carried by the $50$ founders. The effective population size during the bottleneck was very small ( $N_e \approx 50$ ), causing rapid genetic drift that fixed some alleles and eliminated others. The recovered population has reduced heterozygosity and fewer alleles (allelic richness) compared to the pre-bottleneck population.

This is observed in cheetahs (Acinonyx jubatus), which experienced a severe bottleneck approximately $10\,000$ years ago. Modern cheetahs have extremely low genetic diversity and are susceptible to disease and reproductive abnormalities.

(c) A rare disease allele ( $d$ ) that was at very low frequency in the original population may, by chance, be over-represented among the bottleneck survivors. Genetic drift in the small bottleneck population can cause large, random changes in allele frequency. If several of the $50$ survivors happened to carry the $d$ allele, its frequency would increase substantially, leading to a higher disease incidence in the recovered population. This is an example of the founder effect.

Additional Common Pitfalls

Confusing Lamarckian and Darwinian evolution: Lamarck proposed inheritance of acquired characteristics (e.g., giraffes stretch their necks and pass longer necks to offspring); Darwin proposed natural selection acting on random variation. Only Darwinian evolution is supported by evidence.
Assuming evolution has a direction or goal: evolution is not progressive or teleological; it is the non-random survival of randomly generated variants. "More evolved" is not a meaningful concept.
Confusing homology with analogy: homologous structures share a common ancestry (e.g., bat wing and human arm); analogous structures have similar function but different evolutionary origins (e.g., bat wing and insect wing). Convergent evolution produces analogy, not homology.
Misapplying Hardy-Weinberg: H-W assumes no selection, no mutation, no migration, random mating, large population, and no genetic drift. Deviation from H-W indicates that one or more of these conditions is violated, but it does not identify which one.
Assuming natural selection always produces optimal adaptations: natural selection produces adaptations that are "good enough" for the current environment, constrained by historical legacy (e.g., the blind spot in the vertebrate eye), genetic correlations, and physical/chemical constraints.
Confusing genetic drift and natural selection: drift is random change in allele frequency (stronger in small populations); selection is non-random change (favours beneficial alleles).

Additional Exam-Style Problems with Full Solutions

Problem 6: Extended Response -- Evidence for Evolution

Describe and explain four types of evidence that support the theory of evolution by natural selection. For each type, provide a specific example and explain how it supports common descent. (a) Palaeontological evidence, (b) anatomical evidence, (c) molecular evidence, (d) biogeographical evidence.

Answer 6

(a) Palaeontological evidence (fossil record): Fossils show a sequence of progressively changing forms over geological time, documenting the evolution of major groups. Example: the evolution of the horse (Equidae) from Hyracotherium (Eocene, $55\;\mathrm{Ma}$ , dog-sized, four-toed) through Mesohippus, Merychippus, to modern Equus (toe reduction from 4 to 1, increase in body size, changes in tooth morphology for grass-eating). Transitional fossils (e.g., Tiktaalik, showing features intermediate between fish and tetrapods) demonstrate that major evolutionary transitions occurred through gradual modification of existing structures. The fossil record also shows patterns of mass extinction followed by adaptive radiation of surviving groups.

(b) Anatomical evidence (comparative anatomy): Homologous structures (same evolutionary origin, potentially different function) indicate common ancestry. Example: the pentadactyl limb (five-digit limb) in vertebrates -- the human arm, whale flipper, bat wing, and cat leg all have the same basic bone structure (humerus, radius, ulna, carpals, metacarpals, phalanges) despite different functions (manipulation, swimming, flying, walking). Vestigial structures (remnants of structures that had a function in ancestors but are reduced or non-functional in descendants) also support evolution: the human appendix (remnant of a larger caecum in herbivorous ancestors), pelvic bones in whales, wings in flightless birds, and eyes in cave-dwelling fish.

(c) Molecular evidence: All organisms use the same genetic code (DNA/RNA, universal codons), the same basic biochemical pathways (glycolysis, Krebs cycle), and share homologous genes and proteins. The degree of molecular similarity reflects evolutionary relatedness. Example: cytochrome c is a highly conserved protein; humans and chimpanzees have identical cytochrome c sequences, while humans and yeast differ by approximately $45\%$ of amino acids. Molecular clocks (based on neutral mutation rates) allow estimation of divergence times that are consistent with the fossil record. Pseudogenes (non-functional copies of genes that have accumulated mutations) provide a record of gene duplications and losses in evolutionary history.

(d) Biogeographical evidence: The geographic distribution of species reflects evolutionary history. Species on oceanic islands are typically more similar to species on the nearest mainland than to species on other islands (despite similar environments), indicating colonisation from the mainland followed by adaptive radiation. Example: Darwin's finches on the Galapagos Islands are all descended from a single South American finch ancestor and have diversified into species with different beak shapes adapted to different food sources. The unique fauna of Australia (marsupials) reflects its long isolation after the breakup of Gondwana; convergent evolution produced marsupial equivalents of placental mammals (e.g., Tasmanian tiger resembled wolves; marsupial mole resembles placental moles).

Problem 7: Quantitative -- Selection Coefficients and Fitness

In a population of beetles, three colour morphs exist: green (genotype $GG$ ), brown ( $Gg$ ), and black ( $gg$ ). A researcher measures the survival and reproductive success of each genotype:

Genotype	Number born	Number surviving to reproduce	Average offspring per survivor
$GG$	300	240	8
$Gg$	600	420	6
$gg$	100	50	4

(a) Calculate the absolute fitness of each genotype. (b) Calculate the relative fitness of each genotype. (c) Calculate the selection coefficient against each genotype. (d) What type of selection is this? (e) Predict the evolutionary outcome.

Answer 7

(a) Absolute fitness ( $W$ ) = number surviving $\times$ average offspring per survivor:

$GG$ : $240 \times 8 = 1920$
$Gg$ : $420 \times 6 = 2520$
$gg$ : $50 \times 4 = 200$

(b) Relative fitness ( $w$ ) = absolute fitness / maximum absolute fitness: Maximum absolute fitness $= 2520$ ( $Gg$ ).

$GG$ : $1920/2520 = 0.762$
$Gg$ : $2520/2520 = 1.000$
$gg$ : $200/2520 = 0.079$

$GG$ : $s = 1 - 0.762 = 0.238$
$Gg$ : $s = 0$
$gg$ : $s = 1 - 0.079 = 0.921$

(d) This is heterozygote advantage (overdominance, balancing selection). The heterozygote ( $Gg$ ) has the highest fitness, while both homozygotes have lower fitness. This is the genetic basis for the maintenance of genetic polymorphism.

(e) Under heterozygote advantage, both alleles are maintained in the population at a stable equilibrium frequency. The equilibrium frequency of $g$ is: $\hat{q} = \frac{s_{GG}}{s_{GG} + s_{gg}} = \frac{0.238}{0.238 + 0.921} = \frac{0.238}{1.159} = 0.205$

The equilibrium frequency of $G$ is $\hat{p} = 1 - 0.205 = 0.795$ .

At equilibrium, $P(GG) = 0.632$ , $P(Gg) = 0.326$ , $P(gg) = 0.042$ .

This is analogous to sickle-cell allele maintenance in malaria-endemic regions: $HbS$ heterozygotes have malaria resistance, while both homozygotes have reduced fitness (normal homozygotes: susceptible to malaria; $HbS$ homozygotes: sickle-cell anaemia).

Problem 8: Extended Response -- Reproductive Isolation Mechanisms

Two populations of salamanders live on opposite sides of a mountain range. Describe the types of reproductive isolation that could prevent them from interbreeding, classifying each as pre-zygotic or post-zygotic and providing a specific example for each type.

Answer 8

Pre-zygotic barriers (prevent mating or fertilisation):

Habitat isolation: the populations occupy different habitats and do not encounter each other. Example: one population lives in fast-flowing streams, the other in still ponds.
Temporal isolation: the populations breed at different times. Example: one population breeds in spring, the other in autumn.
Behavioural isolation: the populations have different courtship rituals that are not recognised by the other. Example: different pheromone compositions or courtship dances.
Mechanical isolation: physical differences prevent successful mating. Example: incompatible genital morphology (less common in salamanders, more common in insects with "lock-and-key" genitalia).
Gametic isolation: even if mating occurs, the sperm cannot fertilise the egg. Example: species-specific egg jelly layers that only allow conspecific sperm to penetrate.

Post-zygotic barriers (reduce fitness of hybrid offspring):

Hybrid inviability: the zygote does not develop properly or the offspring die early. Example: developmental abnormalities in hybrid embryos due to incompatible gene regulatory networks.
Hybrid sterility: hybrids survive but are sterile. Example: mules (horse $\times$ donkey) are viable but sterile because chromosomes cannot pair properly during meiosis. In salamanders, triploid hybrids would be sterile.
Hybrid breakdown: first-generation hybrids are viable and fertile, but subsequent generations show reduced fitness. Example: F2 hybrids have lower survival or fertility than F1 hybrids.

The mountain range could initially cause allopatric speciation by creating geographic isolation. Over time, genetic drift and natural selection (different environments on each side) would lead to divergence, potentially creating multiple pre-zygotic and post-zygotic barriers. If the populations were to come into secondary contact, these barriers would maintain them as separate species.

Problem 9: Data Analysis -- Allele Frequency Change Over Time

A population of butterflies is monitored for the frequency of the melanic allele ( $M$ ) over 10 years. Industrial pollution darkens tree bark during this period.

Year	$p$ (frequency of $M$ )	$q$ (frequency of $m$ )
0	0.10	0.90
2	0.25	0.75
4	0.45	0.55
6	0.62	0.38
8	0.74	0.26
10	0.80	0.20

(a) Plot $p$ vs time. (b) Calculate the selection coefficient $s$ against the recessive ( $mm$ ) phenotype using the data from year 0 to year 10 (assume the population is in H-W equilibrium each year and selection is against the homozygous recessive). (c) After year 10, pollution is reduced and tree bark lightens. Predict the trend in $p$ over the next 10 years. (d) Explain how this example illustrates natural selection rather than genetic drift.

Answer 9

(a) [Students should plot $p$ on the y-axis ( $0$ to $1$ ) and year on the x-axis ( $0$ to $10$ ). The curve is S-shaped (sigmoidal), characteristic of selection against a recessive phenotype: slow change when $q$ is high (most $M$ alleles are in heterozygotes and protected from selection), accelerating as $q$ decreases, then slowing again as $p$ approaches $1$ .]

(b) For selection against a recessive with fitness $w_{mm} = 1 - s$ and $w_{MM} = w_{Mm} = 1$ : $q' = \frac{q(1 - sq)}{1 - sq^2}$

Using year 0: $q = 0.90$ , $q^2 = 0.81$ . Year 10: $q = 0.20$ . Over 10 years (approximately 5 generations if the generation time is 2 years):

This is complex to solve analytically. Using the approximate formula for one generation: $\Delta q \approx -sq^2(1-q)$

From year 0 to 2: $\Delta q = 0.90 - 0.75 = 0.15$ , $q \approx 0.90$ . $0.15 \approx s \times 0.81 \times 0.10 = 0.081s$ , so $s \approx 1.85$ .

This is greater than 1, which is impossible. This suggests the selection is very strong (near-lethal against $mm$ on dark bark) or that more than one generation occurred in 2 years.

If the generation time is 1 year (10 generations in 10 years): Using the exact formula iteratively with $s = 0.5$ : Gen 0: $q = 0.900$ Gen 1: $q = 0.900(1 - 0.5 \times 0.900)/(1 - 0.5 \times 0.810) = 0.900 \times 0.550/0.595 = 0.832$ Gen 2: $q = 0.832(1 - 0.5 \times 0.832)/(1 - 0.5 \times 0.692) = 0.832 \times 0.584/0.654 = 0.743$ ...

With $s \approx 0.5$ and 10 generations, the trajectory approximately matches the data. The exact value of $s$ depends on the number of generations, but the selection is clearly strong.

(c) After pollution reduction, the selective advantage reverses: light-coloured ( $mm$ ) butterflies are now better camouflaged on light bark, while melanic ( $MM$ , $Mm$ ) butterflies are more visible to predators. The frequency of $M$ would decrease (and $m$ would increase) as selection now acts against the melanic phenotype. This is exactly what was observed with the peppered moth (Biston betularia) in England after the Clean Air Acts of the 1950s--1970s.

(d) This is natural selection rather than drift because: (1) the change is directional and consistent over time (always increasing), which is unlikely to be random; (2) the change is correlated with an environmental change (pollution darkening bark); (3) the mechanism is known (differential predation by birds based on camouflage); (4) the magnitude and direction of change are consistent with the predicted effect of selection. In genetic drift, the direction of change would be random and not consistently correlated with environmental change.

Problem 10: Extended Response -- Convergent Evolution vs Divergent Evolution

(a) Define convergent evolution and divergent evolution, providing two examples of each. (b) Explain why convergent evolution is common in certain environments. (c) Describe how analogous structures can be distinguished from homologous structures using molecular evidence. (d) Explain how the repeated evolution of camera-type eyes in cephalopods (squid) and vertebrates illustrates convergent evolution at both the anatomical and molecular levels.

Answer 10

(a) Convergent evolution: unrelated species independently evolve similar traits due to similar environmental pressures or selective forces. Examples: (1) wings in bats (mammals), birds (aves), and insects (arthropods) -- all used for powered flight but evolved independently from different ancestral structures; (2) streamlined body shape in dolphins (mammals), sharks (fish), and ichthyosaurs (extinct reptiles) -- all adapted for fast swimming in aquatic environments.

Divergent evolution: related species evolve different traits from a common ancestor, usually due to adapting to different environments. Examples: (1) Darwin's finches -- one ancestral finch species diversified into species with different beak shapes adapted to different food sources; (2) the mammalian limb -- the same basic pentadactyl structure has been modified into the horse hoof, bat wing, whale flipper, and human hand.

(b) Convergent evolution is common when: (1) physical laws constrain the solutions available to organisms (e.g., only certain shapes are aerodynamically efficient); (2) similar ecological niches exist in different geographic regions (e.g., desert environments favour succulent plants and nocturnal animals regardless of continent); (3) there are a limited number of optimal solutions to a given problem (e.g., camera-type eyes are highly effective for image formation, so they evolved independently in vertebrates, cephalopods, and some cnidarians).

(c) Anatomically similar structures can be distinguished by comparing their developmental origins and underlying genetic basis:

Homologous structures: develop from the same embryonic tissues, are controlled by homologous genes (e.g., Hox genes), and have similar underlying anatomy despite different functions. Molecular analysis shows they share orthologous genes (genes descended from a common ancestor).
Analogous structures: develop from different embryonic tissues, may be controlled by different genes (or the same genes co-opted independently), and have different internal anatomy despite superficial similarity. Molecular analysis shows they use paralogous genes (gene duplications) or different genes altogether.

(d) Camera-type eyes in cephalopods and vertebrates:

Anatomical convergence: both have a lens, retina, iris, and vitreous humour, and form an inverted image on the retina. However, the developmental origins differ: the vertebrate eye develops as an outgrowth of the brain (the optic cup), while the cephalopod eye develops from an invagination of the skin (ectoderm). The vertebrate retina is "inverted" (photoreceptors face away from the light source, requiring light to pass through other retinal layers), while the cephalopod retina is "everted" (photoreceptors face the light source directly). The cephalopod eye is arguably better designed.
Molecular convergence: both eyes use opsins (light-sensitive proteins) for phototransduction, but the specific opsins differ (vertebrates use ciliary opsins; cephalopods use rhabdomeric opsins). Both use similar phototransduction cascades (G-protein coupled), but the specific proteins are products of different gene families. This demonstrates that natural selection can arrive at similar functional solutions (camera-type eye) using different molecular toolkits -- a remarkable example of convergent evolution at multiple levels.

Genetics and inheritance patterns: Review ./genetics for Mendelian genetics, meiosis, and chromosome behaviour.
DNA, mutations, and molecular biology: Review ./molecular-biology for mutation types and DNA structure.
Advanced genetics and genetic engineering: Review ./genetics-advanced for DNA technology, PCR, and genetic modification.
Ecology and populations: Review ./ecology for population dynamics, communities, and biogeography.
Immunology and pathogen evolution: Review ./immunology for coevolution of hosts and pathogens, antigenic variation.

Supplementary: Molecular Clocks and Phylogenetics (HL Extension)

The Molecular Clock Concept

The molecular clock hypothesis, proposed by Zuckerkandl and Pauling (1962), states that the rate of molecular evolution (nucleotide or amino acid substitution) is approximately constant over time and among lineages. If the rate is known, the amount of molecular divergence between two species can be used to estimate the time since they shared a common ancestor.

Calibration: the clock must be calibrated using known divergence times from the fossil record. For example, if two species diverged $10\;\mathrm{Ma}$ and differ by $5\%$ in a given gene, the substitution rate is $0.5\%$ per million years.

Types of Molecular Markers

Different genes and genomic regions evolve at different rates, making them suitable for different timescales of evolutionary analysis:

Marker	Substitution rate	Time depth	Application
rRNA (18S, 28S)	Very slow	$>500\;\mathrm{Ma}$	Deep phylogeny (kingdoms, phyla)
Mitochondrial COI ("DNA barcode")	Moderate	$1$ -- $100\;\mathrm{Ma}$	Species identification, recent divergences
Mitochondrial control region	Fast	$0.01$ -- $1\;\mathrm{Ma}$	Population genetics, intraspecific variation
Microsatellites (STRs)	Very fast	$<0.01\;\mathrm{Ma}$	Individual identification, parentage
Pseudogenes	Fast (no functional constraint)	Variable	Gene family evolution

Assumptions and Limitations of the Molecular Clock

Rate constancy: the clock assumes a constant substitution rate. In reality, rates vary due to:
- Generation time (shorter generation = more replications per year = faster molecular evolution).
- Metabolic rate (higher metabolism = more oxidative damage = faster mutation rate).
- Body size (smaller organisms tend to have faster rates).
- DNA repair efficiency.
- Functional constraints (conserved genes evolve more slowly than non-functional sequences).
Neutral theory: the clock works best for neutral mutations (not subject to natural selection). Positively selected sites evolve faster (adaptive evolution); negatively selected (purifying selection) sites evolve more slowly.
Saturation: at large evolutionary distances, multiple substitutions at the same site can obscure the true number of changes (multiple hits). This leads to underestimation of divergence.

Phylogenetic Tree Construction Methods

Distance-based methods (e.g., neighbour-joining):

Calculate a pairwise distance matrix (number of substitutions between each pair of sequences).
Build a tree that minimises the total tree length (sum of all branch lengths).
Fast and widely used but discard information about specific substitutions.

Character-based methods (e.g., maximum parsimony, maximum likelihood, Bayesian inference):

Maximum parsimony: finds the tree requiring the fewest evolutionary changes. Simple but sensitive to long-branch attraction (long branches may be artefactually grouped together).
Maximum likelihood: finds the tree that maximises the probability of observing the data given a specific model of evolution (e.g., Jukes-Cantor, HKY85, GTR). Computationally intensive but statistically rigorous.
Bayesian inference: calculates the posterior probability of each tree given the data and a prior. Provides probability distributions for tree topology and branch lengths. Used in programs like MrBayes and BEAST.

Interpreting Phylogenetic Trees

The root represents the common ancestor of all taxa in the tree.
Branch lengths can represent the number of substitutions (in a cladogram, all branches are equal length) or the amount of evolutionary change (in a phylogram, branch lengths are proportional to the number of substitutions).
Monophyletic group (clade): an ancestor and ALL of its descendants. All members share a most recent common ancestor not shared with any member outside the group.
Paraphyletic group: an ancestor and SOME (but not all) of its descendants. Example: "reptiles" (excludes birds, which are descended from reptiles).
Polyphyletic group: taxa that do not share a most recent common ancestor within the group. Example: "warm-blooded animals" (birds and mammals, whose common ancestor was cold-blooded).

Worked Example: Molecular Clock Dating with Multiple Loci

Three species of finch from the Galapagos Islands are compared at two genes:

Gene A (slow clock, rate $= 1 \times 10^{-9}$ substitutions/site/year):

Species 1 vs 2: $12$ substitutions per $1000\;\mathrm{bp}$
Species 1 vs 3: $18$ substitutions per $1000\;\mathrm{bp}$
Species 2 vs 3: $6$ substitutions per $1000\;\mathrm{bp}$

Gene B (fast clock, rate $= 5 \times 10^{-9}$ substitutions/site/year):

Species 1 vs 2: $50$ substitutions per $1000\;\mathrm{bp}$
Species 1 vs 3: $80$ substitutions per $1000\;\mathrm{bp}$
Species 2 vs 3: $30$ substitutions per $1000\;\mathrm{bp}$

(a) Construct the most parsimonious tree. (b) Estimate divergence times using each gene separately. (c) Explain why the two genes give different time estimates.

Solution

(a) Gene A: the smallest distance is between species 2 and 3 ( $6/1000$ ). The larger distances are species 1 vs 2 ( $12/1000$ ) and species 1 vs 3 ( $18/1000$ ). Since $12 + 6 = 18$ , the tree is:

     ┌── Species 2 (6 substitutions from node)
─────┤
     └────── Species 3 (12 from node)
         (root)── Species 1

Species 2 and 3 are sister taxa; species 1 is the outgroup.

Gene B: same topology (smallest distance between 2 and 3: $30/1000$ ; $50 + 30 = 80$ ).

Both genes support the same tree topology: (2,3) are sister taxa.

(b) Gene A: rate $= 1 \times 10^{-9}$ subs/site/year.

Species 2--3 divergence: $6/1000 = 0.006$ subs/site. Time $= 0.006 / (2 \times 10^{-9}) = 3 \times 10^6$ years $= 3\;\mathrm{Ma}$ . (Dividing by 2 because substitutions accumulate on both lineages.)
Species 1--(2,3) divergence: $(12 + 6)/2000 = 0.009$ subs/site from the root. Or: the distance from species 1 to the common ancestor of 1,2,3: $(12 - 3)/1000 = 0.009$ (correcting for the 2--3 split). Time $= 0.009 / (2 \times 10^{-9}) = 4.5\;\mathrm{Ma}$ .

Gene B: rate $= 5 \times 10^{-9}$ subs/site/year.

Species 2--3: $30/1000 = 0.030$ . Time $= 0.030 / (2 \times 5 \times 10^{-9}) = 3.0 \times 10^6 = 3.0\;\mathrm{Ma}$ .
Species 1--(2,3): $(50 - 15)/1000 = 0.035$ . Time $= 0.035 / (2 \times 5 \times 10^{-9}) = 3.5\;\mathrm{Ma}$ .

(c) The two genes give slightly different divergence times for the deeper node (Gene A: $4.5\;\mathrm{Ma}$ ; Gene B: $3.5\;\mathrm{Ma}$ ). This discrepancy is due to:

Rate variation between genes (the molecular clock is not perfectly constant).
Stochastic variation (random nature of mutation).
Different functional constraints on the two genes (Gene B may have slightly relaxed constraints, leading to a faster effective rate in one lineage).
Multiple hits at the same site (saturation), which is more problematic for the faster-evolving Gene B.

Using multiple loci and methods that account for rate variation (relaxed molecular clock models) can provide more accurate estimates.

Supplementary: Evolutionary Developmental Biology (Evo-Devo) (HL Extension)

What is Evo-Devo?

Evolutionary developmental biology (evo-devo) studies how changes in developmental processes generate evolutionary change. The central insight is that evolution acts not only on genes but on the genetic regulatory networks that control development.

Key Concepts

Modularity: developmental processes are modular -- different body parts (segments, organs) develop relatively independently under the control of specific sets of genes. This allows changes in one module without affecting others, facilitating evolutionary innovation.

Heterotopy: change in the location of a developmental process. Example: the expression of legs instead of antennae in Drosophila (Antennapedia mutation) is a heterotopic change.

Heterochrony: change in the timing of a developmental process. Example: paedomorphosis (retention of juvenile features in adults) -- the axolotl retains larval gills and aquatic lifestyle as an adult. Hypermorphosis (extended development) produces larger or more complex features.

Gene duplication: duplication of developmental genes (e.g., Hox clusters, PAX genes) allows one copy to maintain the original function while the other is free to evolve new functions (neofunctionalisation) or subdivide the original function (subfunctionalisation). The four Hox clusters in vertebrates arose from two rounds of whole-genome duplication early in vertebrate evolution.

Examples of Evo-Devo

Stickleback pelvic reduction: freshwater stickleback populations have lost their pelvic spines. This is caused by deletion of a pelvic-specific enhancer of the Pitx1 gene. The Pitx1 gene still functions normally in other tissues (head, jaw), but the loss of the pelvic enhancer prevents expression in the pelvic region. This illustrates how changes in regulatory elements (not coding sequences) can produce morphological evolution.

Darwin's finches: variation in beak shape and size is controlled in part by the expression levels of BMP4 (broad, deep beaks) and Calmodulin (long, pointed beaks). Small changes in the timing and level of expression of these genes during embryonic development produce the diverse beak morphologies observed across the Galapagos finch species.

Sonic hedgehog (Shh) and limb development: the zone of polarising activity (ZPA) at the posterior of the limb bud produces Shh protein, which establishes the anterior-posterior axis of the limb. The amount of Shh produced determines the number and type of digits. Mutations affecting Shh expression can cause polydactyly (extra digits) or loss of digits.

Snake limblessness: snakes lost their limbs during evolution. The Shh and HoxD expression domains in the limb buds of snake embryos are reduced or absent compared to limbed reptiles. Changes in the regulatory DNA controlling these genes (not the genes themselves) are responsible for limb loss.

Worked Example: Calculating Morphological Change Rate

A fossil sequence shows that the average body size of a mammal lineage increased from $5\;\mathrm{kg}$ to $50\;\mathrm{kg}$ over $10$ million years. (a) Calculate the rate of body size increase in $\mathrm{mg/year}$ . (b) Using the equation for exponential growth of body size ( $M_t = M_0 e^{rt}$ ), calculate the rate constant $r$ . (c) If Cope's rule (the tendency for lineages to increase in body size over time) applies, predict the body size after $20$ million years from the original ancestor.

Solution

(a) Rate of change: $\frac{50\;\mathrm{kg} - 5\;\mathrm{kg}}{10 \times 10^6\;\mathrm{years}} = \frac{45}{10^7} = 4.5 \times 10^{-6}\;\mathrm{kg/year} = 4.5\;\mathrm{mg/year}$ .

(b) $M_t = M_0 e^{rt}$ : $50 = 5 e^{10r}$ . $e^{10r} = 10$ , so $10r = \ln(10) = 2.303$ , $r = 0.230\;\mathrm{Ma}^{-1}$ .

Cope's rule predicts continued body size increase, but in reality, body size eventually plateaus due to physiological and ecological constraints (larger organisms require more food, have longer generation times, and face biomechanical limitations). Many lineages do not follow Cope's rule; body size can decrease (island dwarfism) or fluctuate depending on environmental conditions.

1. Evidence for Evolution​

Fossil Evidence​

Anatomical Evidence​

Molecular Evidence​

Embryological Evidence​

Biogeographical Evidence​

2. Natural Selection in Detail​

Types of Natural Selection​

Directional Selection​

Stabilising Selection​

Disruptive Selection​

Sexual Selection​

Frequency-Dependent Selection​

3. Speciation​

Allopatric Speciation​

Sympatric Speciation​

Parapatric Speciation​

Reproductive Isolating Mechanisms​

4. Adaptive Radiation and Convergent Evolution​

Adaptive Radiation​

Convergent Evolution​

5. Gradualism vs Punctuated Equilibrium​

Phyletic Gradualism (Darwinian Gradualism)​

Punctuated Equilibrium (Gould and Eldredge, 1972)​

6. Hardy-Weinberg Equilibrium (Extended)​

The Principle​

Testing Hardy-Weinberg with Multiple Alleles​

Effects of Evolutionary Forces on Hardy-Weinberg​

Mutation​

Migration (Gene Flow)​

Genetic Drift​

Non-Random Mating​

Natural Selection​

Fitness and Selection Coefficients​

7. Phylogenetic Trees​

Construction​

Types of Trees​

Interpreting Phylogenetic Trees​

Molecular Phylogenetics​

8. Antibiotic Resistance as Evolution​

Common Pitfalls​

Practice Problems​

Worked Examples​

Common Pitfalls (Expanded)​

Exam-Style Problems​

If You Get These Wrong, Revise:​

Additional Worked Examples​

Additional Common Pitfalls​

Additional Exam-Style Problems with Full Solutions​

Cross-References to Related Topics​

Supplementary: Molecular Clocks and Phylogenetics (HL Extension)​

The Molecular Clock Concept​

Types of Molecular Markers​

Assumptions and Limitations of the Molecular Clock​

Phylogenetic Tree Construction Methods​

Interpreting Phylogenetic Trees​

Worked Example: Molecular Clock Dating with Multiple Loci​

Supplementary: Evolutionary Developmental Biology (Evo-Devo) (HL Extension)​

What is Evo-Devo?​

Key Concepts​

Examples of Evo-Devo​

Worked Example: Calculating Morphological Change Rate​

1. Evidence for Evolution

Fossil Evidence

Anatomical Evidence

Molecular Evidence

Embryological Evidence

Biogeographical Evidence

2. Natural Selection in Detail

Types of Natural Selection

Directional Selection

Stabilising Selection

Disruptive Selection

Sexual Selection

Frequency-Dependent Selection

3. Speciation

Allopatric Speciation

Sympatric Speciation

Parapatric Speciation

Reproductive Isolating Mechanisms

4. Adaptive Radiation and Convergent Evolution

Adaptive Radiation

Convergent Evolution

5. Gradualism vs Punctuated Equilibrium

Phyletic Gradualism (Darwinian Gradualism)

Punctuated Equilibrium (Gould and Eldredge, 1972)

6. Hardy-Weinberg Equilibrium (Extended)

The Principle

Testing Hardy-Weinberg with Multiple Alleles

Effects of Evolutionary Forces on Hardy-Weinberg

Mutation

Migration (Gene Flow)

Genetic Drift

Non-Random Mating

Natural Selection

Fitness and Selection Coefficients

7. Phylogenetic Trees

Construction

Types of Trees

Interpreting Phylogenetic Trees

Molecular Phylogenetics

8. Antibiotic Resistance as Evolution

Common Pitfalls

Practice Problems

Worked Examples

Common Pitfalls (Expanded)

Exam-Style Problems

If You Get These Wrong, Revise:

Additional Worked Examples

Additional Common Pitfalls

Additional Exam-Style Problems with Full Solutions

Cross-References to Related Topics

Supplementary: Molecular Clocks and Phylogenetics (HL Extension)

The Molecular Clock Concept

Types of Molecular Markers

Assumptions and Limitations of the Molecular Clock

Phylogenetic Tree Construction Methods

Interpreting Phylogenetic Trees

Worked Example: Molecular Clock Dating with Multiple Loci

Supplementary: Evolutionary Developmental Biology (Evo-Devo) (HL Extension)

What is Evo-Devo?

Key Concepts

Examples of Evo-Devo

Worked Example: Calculating Morphological Change Rate