Evolution I: Evolutionary Thought, Selection, Life History, and Population and Quantitative Genetics
1. Lamarckism, Darwinism, the Modern Synthesis and the types of selection
Lamarck (Philosophie zoologique, 1809) proposed that organisms change through use and disuse of organs and that characters acquired in a lifetime are inherited; the giraffe’s neck lengthened by stretching. Darwin (On the Origin of Species, 1859), with Wallace, proposed natural selection: individuals vary, much of the variation is heritable, more offspring are produced than survive, and those whose heritable traits suit the environment leave more descendants. Darwin lacked a mechanism of inheritance; the Modern Synthesis of the 1930s–40s (Fisher, Haldane, Wright, Dobzhansky, Mayr, Simpson, Huxley) united Mendelian genetics with selection, showing that continuous variation arises from many Mendelian loci and that evolution is a change in allele frequencies driven by mutation, selection, drift and gene flow.
Fitness is reproductive success: absolute fitness is the expected number of offspring, relative fitness w is fitness divided by that of the best genotype (or by the mean), and the selection coefficient s = 1 − w measures the disadvantage. An adaptation is a trait shaped by selection for its current function; a trait that arose for one function and was later co-opted for another (feathers, perhaps first for insulation) is an exaptation. Not every trait is an adaptation — some are by-products of other traits or of history.
| Mode | Favours | Effect on mean and variance | Example |
|---|---|---|---|
| Stabilising | intermediate phenotypes | mean unchanged, variance reduced | human birth weight |
| Directional | one extreme | mean shifts, variance may fall | larger beaks in Galápagos medium ground finches after the 1977 drought |
| Disruptive | both extremes | mean may stay, variance increases; can become bimodal | two beak sizes in the African black-bellied seedcracker, matched to hard and soft seeds |
2. Life history theory: allocation, trade-offs, r/K and semelparity
An organism has a finite budget of energy and time to divide among growth, maintenance, survival and reproduction, so allocation to one function comes at the expense of another. These trade-offs shape life histories: current reproduction against future survival (the cost of reproduction), offspring number against offspring size (the Smith–Fretwell trade-off), and early against late reproduction. Lack’s clutch size hypothesis says birds lay the number of eggs that maximises the number of fledglings raised, which is often fewer than the maximum they could lay, because very large broods are underfed.
| Feature | r-selected | K-selected |
|---|---|---|
| Environment | variable, unpredictable, often disturbed | stable, crowded, near carrying capacity |
| Offspring | many, small, little parental care | few, large, extensive parental care |
| Maturity and lifespan | early maturity, short life | late maturity, long life |
| Survivorship | usually Type III | usually Type I or II |
| Examples | weedy annual plants, many insects | elephants, great apes, large trees |
Semelparous organisms reproduce once and die — Pacific salmon, many annual plants, bamboos that flower gregariously after decades, Agave; iteroparous organisms reproduce repeatedly — most vertebrates and perennial plants. Semelparity is favoured when adult survival is low or uncertain relative to juvenile survival, or when a large one-off investment gives disproportionately high returns (an Agave’s tall inflorescence attracting pollinators). Cole’s paradox noted that a semelparous annual needs only one more offspring per year than an iteroparous perennial to match it; the resolution is that juvenile survival is usually far lower than adult survival. The r/K scheme has largely given way to demographic life-history theory, in which age-specific mortality drives the evolution of these traits.
3. Variation, Mendelian genetics, Hardy–Weinberg, population structure and gene flow
Genetic variation originates in mutation — point mutations, insertions and deletions, duplications, chromosomal rearrangements — and is reshuffled each generation by recombination, independent assortment and random fertilisation. Mendel’s laws of segregation (a monohybrid F₂ of 3:1, genotypes 1:2:1) and independent assortment (a dihybrid F₂ of 9:3:3:1) hold for unlinked loci. Epistasis, an interaction between loci in which one gene masks or modifies another, changes the dihybrid ratio: recessive epistasis gives 9:3:4 (coat colour in Labrador retrievers), duplicate recessive (complementary) epistasis 9:7, dominant epistasis 12:3:1 and duplicate dominant 15:1.
The Hardy–Weinberg principle: in a large, randomly mating population with no mutation, selection or migration, allele frequencies p and q (p + q = 1) stay constant and genotype frequencies reach p², 2pq and q² after one generation of random mating. Two uses dominate questions. Counting alleles from genotypes: with 360 AA, 480 Aa and 160 aa, p = (2 × 360 + 480)/2000 = 0.6, and the expected heterozygotes 2 × 0.6 × 0.4 × 1000 = 480 match the observed. Inferring carriers from a recessive condition: if 1 in 400 people shows a recessive trait, q² = 0.0025, q = 0.05, p = 0.95 and the carrier frequency is 2pq = 0.095, nearly 1 in 10 — far more carriers than affected individuals.
Population genetic structure: a panmictic population mates at random throughout; real populations are subdivided. Wright’s F_ST = (H_T − H_S)/H_T measures the fraction of total heterozygosity due to differences between subpopulations, where H_S is the mean expected heterozygosity within subpopulations and H_T the expected heterozygosity of the pooled population. With H_S = 0.30 and H_T = 0.40, F_ST = 0.25 — great differentiation by Wright’s rough guide (0–0.05 little, 0.05–0.15 moderate, 0.15–0.25 great, above 0.25 very great). Gene flow homogenises: in the island model at equilibrium, F_ST ≈ 1/(4Nm + 1), where Nm is the number of migrants per generation, so about one migrant per generation (F_ST ≈ 0.2) is enough to prevent subpopulations drifting to fixation for different alleles. Migration changes allele frequency by Δp = m(p_m − p).
4. Genetic drift and effective population size
Genetic drift is random change in allele frequency from generation to generation because only a sample of gametes forms the next generation. Its effects are stronger in small populations: allele frequencies wander, alleles are lost or fixed, and heterozygosity declines. A new neutral mutation in a diploid population of N has a probability 1/(2N) of eventually being fixed, and every neutral allele’s probability of fixation equals its current frequency. The founder effect (a new population started by a few individuals) and bottlenecks (a drastic temporary reduction) are episodes of intense drift; the northern elephant seal, hunted to a few tens of animals in the 1890s, still shows very low genetic variation.
The effective population size Nₑ is the size of an idealised population that would drift at the same rate; it is usually much smaller than the census size. Three standard corrections: an unequal sex ratio, Nₑ = 4N_mN_f/(N_m + N_f) — 10 males and 40 females give Nₑ = 1600/50 = 32, not 50; fluctuating size, where Nₑ is the harmonic mean of the generation sizes — 1000, 10 and 1000 give 3/(0.001 + 0.1 + 0.001) = 29.4, dominated by the bottleneck; and variance in family size, which lowers Nₑ further. Heterozygosity is lost at a rate of 1/(2Nₑ) per generation: H_t = H₀(1 − 1/(2Nₑ))^t. With Nₑ = 25 the loss is 2 % per generation and after 10 generations H₁₀/H₀ = 0.98¹⁰ = 0.817.
5. Selection at one locus with two alleles
Give the genotypes AA, Aa and aa relative fitnesses w₁₁, w₁₂ and w₂₂. After selection the frequency of a is q′ = (pq w₁₂ + q² w₂₂)/w̄, where the mean fitness is w̄ = p²w₁₁ + 2pq w₁₂ + q²w₂₂. For selection against a recessive (w₁₁ = w₁₂ = 1, w₂₂ = 1 − s) this simplifies to q′ = q(1 − sq)/(1 − sq²) and Δq = −sq²(1 − q)/(1 − sq²). Worked example, q = 0.4 and s = 0.5: w̄ = 1 − 0.5 × 0.16 = 0.92 and q′ = 0.4 × 0.8/0.92 = 0.32/0.92 = 0.348. Selection against a rare recessive is very slow, because rare recessive alleles hide in heterozygotes where selection cannot see them.
| Fitness pattern | Outcome | Key result |
|---|---|---|
| Directional (one homozygote best) | the favoured allele goes to fixation | slowest when the favoured allele is recessive and rare |
| Overdominance (heterozygote best: w = 1 − s₁, 1, 1 − s₂) | stable polymorphism | q̂ = s₁/(s₁ + s₂) for the allele whose homozygote has fitness 1 − s₂ |
| Underdominance (heterozygote worst) | unstable equilibrium; one allele fixed | outcome depends on starting frequency |
| Mutation–selection balance (recessive) | deleterious allele kept at low frequency | q̂ ≈ √(μ/s); for μ = 10⁻⁶ and a lethal (s = 1), q̂ = 0.001 |
The sickle-cell allele Hb^S in malarial regions is the classic case of overdominance: the heterozygote is protected against Plasmodium falciparum malaria, the Hb^A homozygote is susceptible and the Hb^S homozygote suffers sickle-cell disease. If w(AA) = 0.9 (s₁ = 0.1) and w(SS) = 0.2 (s₂ = 0.8), the equilibrium frequency of S is 0.1/0.9 = 0.11, and it persists at that frequency despite being nearly lethal in homozygotes. Frequency-dependent selection can also maintain polymorphism: when rare types are favoured (negative frequency dependence), as in self-incompatibility alleles or scale-eating cichlids with left- and right-bent mouths, each allele rises when rare.
6. Quantitative genetics, plasticity and epigenetic inheritance
Polygenic traits — height, body mass, yield — are controlled by many loci of small effect plus the environment, and vary continuously. The phenotypic variance partitions as V_P = V_G + V_E (plus a genotype-by-environment interaction V_G×E and covariance where they exist), and the genetic variance as V_G = V_A + V_D + V_I: additive, dominance and interaction (epistatic) variance. Broad-sense heritability H² = V_G/V_P; narrow-sense heritability h² = V_A/V_P, the part that makes offspring resemble their parents and the part selection acts on. It is estimated from the slope of offspring on mid-parent values (the slope equals h²), from twins or from the response to selection.
The breeder’s equation R = h²S predicts the response to selection R (the change in the mean between generations) from the selection differential S (the mean of the selected parents minus the population mean). If the population mean is 50, the selected parents average 60 and their offspring average 54, then S = 10, R = 4 and the realised heritability h² = R/S = 0.4. Heritability is a property of a population in an environment, not of a trait: it can differ between populations, and a high heritability within groups says nothing about the cause of a difference between groups.
Phenotypic plasticity is the ability of one genotype to produce different phenotypes in different environments, described by its norm of reaction (phenotype plotted against environment). When the norms of different genotypes are not parallel — they cross or diverge — there is a genotype-by-environment interaction: the best genotype in one environment is not the best in another. Examples are the helmet induced in Daphnia by predator chemicals, temperature-dependent sex determination in many turtles and crocodilians, and the leaf forms of amphibious plants. Epigenetic inheritance passes on states of gene expression without a change in DNA sequence — DNA methylation, histone modification and small RNAs. Genomic imprinting in mammals and paramutation in maize are established examples; the extent and stability of transgenerational epigenetic effects in animals remain actively studied.
Key takeaways
- The Modern Synthesis made evolution a change in allele frequencies by mutation, selection, drift and gene flow; stabilising selection cuts variance, directional shifts the mean, disruptive favours both extremes.
- Life histories are trade-offs: r-selected species breed early and often with small offspring, K-selected late with few large ones; semelparity pays when adult survival is poor.
- Hardy–Weinberg: p² + 2pq + q² = 1; from q², carriers are 2pq. F_ST = (H_T − H_S)/H_T, and about one migrant per generation prevents fixation of different alleles.
- Nₑ = 4N_mN_f/(N_m + N_f), or the harmonic mean over generations; heterozygosity falls by 1/(2Nₑ) per generation, and a new neutral mutation fixes with probability 1/(2N).
- Against a recessive, q′ = q(1 − sq)/(1 − sq²); overdominance holds q̂ = s₁/(s₁ + s₂); the breeder’s equation R = h²S uses narrow-sense heritability V_A/V_P.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The idea that characters acquired during an organism’s lifetime through use and disuse are passed to its offspring is central to
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Answer: A — Lamarckism
Inheritance of acquired characters through use and disuse is Lamarck’s mechanism. Darwin’s is selection among heritable variants, and the Modern Synthesis joined that to Mendelian genetics; the neutral theory concerns molecular substitutions by drift.Infants of intermediate birth weight have the highest survival, and very light and very heavy babies the lowest. Over generations this pattern of selection is expected to
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Answer: A — keep the mean roughly constant and reduce the variance
Favouring intermediates against both extremes is stabilising selection, which trims the tails and so reduces variance without moving the mean. Directional selection moves the mean; disruptive selection widens the distribution. Being polygenic does not stop a trait responding to selection.Which traits are usually associated with K-selected species?
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Answer: A — Delayed maturity; B — Few, large offspring with extensive parental care; C — Populations held near carrying capacity
K-selection operates in crowded, stable environments near K, favouring late maturity and a few well-provisioned offspring. A Type III curve, with heavy juvenile mortality, is typical of r-selected species producing many small offspring.Life history theory predicts that semelparity is most likely to be favoured over iteroparity when
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Answer: A — adult survival from one breeding season to the next is low relative to juvenile survival
If an adult is unlikely to survive to breed again, holding reserves back for future reproduction gains little, and putting everything into one bout pays. High adult survival with poor juvenile survival is the resolution of Cole’s paradox in favour of iteroparity; costless reproduction removes the trade-off altogether.In a sample of 1000 individuals there are 360 AA, 480 Aa and 160 aa. The frequency of allele A, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 0.6
Count alleles: A = 2 × 360 + 480 = 1200 of 2000, so p = 0.6. Taking the AA frequency 0.36 as p forgets the heterozygotes, and √0.36 = 0.6 happens to agree only because the sample is in Hardy–Weinberg proportions.A recessive condition affects 1 in 400 newborns in a randomly mating population. The expected frequency of heterozygous carriers, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.095
q² = 1/400 = 0.0025, q = 0.05, p = 0.95, and 2pq = 2 × 0.95 × 0.05 = 0.095. Answering 0.05 gives the allele frequency, and 0.0025 the affected frequency; carriers outnumber affected individuals by about 38 to 1.Allele a has frequency q = 0.4. Genotypes AA and Aa have fitness 1 and aa has fitness 0.5. After one generation of selection (random mating before selection), the frequency of a, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.348
w̄ = 1 − sq² = 1 − 0.5 × 0.16 = 0.92. q′ = (pq + q²(1 − s))/w̄ = (0.24 + 0.08)/0.92 = 0.32/0.92 = 0.348, so Δq = −0.052. Forgetting to divide by w̄ gives 0.32.At a locus with heterozygote advantage, the fitnesses of AA, AS and SS are 0.9, 1 and 0.2. The equilibrium frequency of allele S, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.11
With s₁ = 0.1 against AA and s₂ = 0.8 against SS, q̂(S) = s₁/(s₁ + s₂) = 0.1/0.9 = 0.11. Writing 0.8/0.9 = 0.89 gives the frequency of A instead: each allele’s equilibrium is set by the selection against the other homozygote.A captive breeding population has 10 breeding males and 40 breeding females. Its effective population size is ____.
Numerical answer — type the value.
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Answer: 32
Nₑ = 4N_mN_f/(N_m + N_f) = 4 × 10 × 40/50 = 32. The census size 50 overstates Nₑ because half of all genes in each generation come through only ten males.A population has 1000, 10 and 1000 breeding individuals in three successive generations. Its effective size over the three generations, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 29.4
Nₑ is the harmonic mean: 3/(1/1000 + 1/10 + 1/1000) = 3/0.102 = 29.4. The arithmetic mean, 670, ignores that the single bottleneck generation dominates the loss of variation.An isolated population has a constant effective size Nₑ = 25. The fraction of its initial heterozygosity remaining after 10 generations of drift, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.817
Each generation loses 1/(2Nₑ) = 1/50 = 0.02, so H₁₀/H₀ = (1 − 0.02)¹⁰ = 0.98¹⁰ = 0.817. Subtracting 10 × 0.02 gives 0.8, which treats a compounding loss as additive; using 1/Nₑ gives 0.96¹⁰ = 0.665.The mean expected heterozygosity within subpopulations is H_S = 0.30, and the expected heterozygosity of the pooled population is H_T = 0.40. Wright’s F_ST, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.25
F_ST = (H_T − H_S)/H_T = (0.40 − 0.30)/0.40 = 0.25. Dividing by H_S gives 0.33, a common slip; a quarter of the total heterozygosity is due to differences between subpopulations.In a selection experiment, the population mean for a trait is 50, the selected parents have a mean of 60 and their offspring have a mean of 54. The realised heritability, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 0.4
S = 60 − 50 = 10 and R = 54 − 50 = 4, so h² = R/S = 0.4. Using the offspring mean over the parent mean, 54/60 = 0.9, confuses a ratio of means with the ratio of deviations from the population mean.Which statements about heritability are correct?
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Answer: A — Narrow-sense heritability is V_A/V_P and determines the response to selection; B — The slope of offspring values regressed on mid-parent values estimates h²; C — Heritability can differ between populations and environments for the same trait
h² = V_A/V_P enters R = h²S, and the offspring–mid-parent slope estimates it. Being a ratio of variances in a population, it changes with the population and environment. It describes variation among individuals, not the make-up of one individual’s phenotype, so the last statement misreads it.A dihybrid cross for coat colour gives an F₂ of 9 black : 3 brown : 4 yellow. The interaction is best described as
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Answer: A — recessive epistasis
9:3:4 arises when homozygosity for the recessive allele at one locus (ee) masks the other locus, merging 3 + 1 into 4 — the Labrador retriever pattern. Dominant epistasis gives 12:3:1 and complementary action 9:7; a single locus with incomplete dominance gives 1:2:1.Two genotypes of a plant are grown at low and high nitrogen. Genotype G1 is taller than G2 at low nitrogen, but G2 is taller than G1 at high nitrogen. Which conclusions follow?
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Answer: A — There is a genotype-by-environment interaction for height; B — The norms of reaction of G1 and G2 cross; C — Height in these genotypes is phenotypically plastic
The rank of the genotypes reverses between environments, so the reaction norms cross — the defining sign of G × E. Each genotype’s height changes with nitrogen, which is plasticity. The genotypes differ in each environment, so genetic variance is present, not zero.A fully recessive lethal allele arises by mutation at a rate μ = 1 × 10⁻⁶ per generation. Its equilibrium frequency under mutation–selection balance, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.001
For a recessive, q̂ ≈ √(μ/s) = √(10⁻⁶/1) = 10⁻³ = 0.001. For a dominant or additive deleterious allele the balance is q̂ ≈ μ/s (here 10⁻⁶), much lower, because selection sees it in heterozygotes.