Biology · Biodiversity, Global Change & Conservation
Extinction risk, small populations and genetic consequences
If two individuals independently have a one-half chance of surviving an interval, the mean surviving count is one. Yet one of the four equally likely complete outcomes has no survivor. Expected abundance and the probability of losing every member answer different questions, and conservation risk also depends on what the population boundary includes.
Begin with the biological levels, species identity and ordinary fractions introduced in the earlier biology lessons. Population Ecology & Life-History Patterns helps with demographic accounts, expected counts and conditional probabilities. Ecosystems, Energy Flow & Biogeochemical Cycles supplies the distinction between a process, a stock and an ecosystem boundary, while Species, Speciation & Macroevolution helps with the meaning and limits of species identities. The copy-pool probability examples can be read as complete sets of alternatives before learning binomial notation; return to the genetics lessons for inheritance, drift and relatedness.
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Reason about population-loss risk and genetic variation with explicit boundaries and model conditions, without treating abundance or a risk category as a guaranteed fate.
Extinction risk has a boundary, horizon and conditions
Global species extinction means that no living member of the species remains. Local population loss, often called extirpation, means disappearance from a specified area while the species can remain elsewhere. Loss of an allele is another event: a genetic variant disappears from the included copy pool. A declining local count cannot automatically be renamed a global extinction or an allele-frequency measurement.
Risk concerns a possible event under defined conditions and over a stated horizon. Two populations with the same abundance can differ in reproductive contribution, isolation, trends, habitat and exposure to shared threats. Natural rarity and rapid decline are also distinct: a historically restricted population may face different conditions from a once-abundant population undergoing sustained losses. Neither description removes the need to assess its actual pressures.
An extinction-risk assessment summarises evidence such as population trend, distribution and threats. IUCN’s threatened categories include Vulnerable, Endangered and Critically Endangered. They are not statements that extinction is certain or a complete ordering of every conservation decision. Data Deficient means that available evidence does not resolve an appropriate risk category; it does not establish that the species is safe. No current species assignment is made here.
Copy sampling, inbreeding and adaptive potential differ
Random genetic drift changes allele frequencies through finite contributions from one generation to the next. In the neutral four-copy example, each next-generation copy is an independent replacement draw from an equal A/a pool. Some outcomes lose A and some lose a; they cancel in the mean A frequency. Variation within the resulting pools nonetheless falls on average. A neutral loss does not show that the lost allele was harmful.
Inbreeding concerns reproduction among related individuals and increased probability of alleles being identical by descent. It can increase homozygosity and expose effects of recessive deleterious variants; it does not create those variants simply by pairing relatives. Its consequences depend on genetic history and context. Observed heterozygosity, allele richness, relatedness and fitness are therefore connected questions rather than interchangeable measurements.
The genetically effective population size compares a real population’s drift or inbreeding behaviour with a specified idealised model. Unequal contributions and other departures from ideal assumptions can make it differ from a census count. Gene flow can add variation and has benefited selected inbred populations, but adaptation, ecological compatibility and possible adverse consequences need assessment. The possibility of genetic rescue is not a guarantee of demographic recovery, and these concepts supply no release or breeding instructions.
Variation can fall while the average frequency stays the same
Read the diagram in words
A-counts 0, 1, 2, 3, 4 have probabilities 1/16, 4/16, 6/16, 4/16, 1/16. Their mean A frequency is 0.5. Pool-diversity values are 0, 0.375, 0.5, 0.375, 0 and their probability-weighted mean is 0.375, below initial 0.5. Either single-allele endpoint occurs with combined probability 2/16 = 0.125.
| Initial four-copy pool | = |
|---|---|
| Initial four-copy pool | A · A · a · a |
| N | 2 |
| 2N | 4 |
| Fraction of copies that are A | 0.5 |
| Probability that independent pool draws differ | 0.5 |
| Next-generation copy positions in the supplied order | Number of A copies | Fraction of copies that are A | Probability that independent pool draws differ | P(A/a) |
|---|---|---|---|---|
| a · a · a · a | 0 | 0 | 0 | 0.0625 |
| a · a · a · A | 1 | 0.25 | 0.375 | 0.0625 |
| a · a · A · a | 1 | 0.25 | 0.375 | 0.0625 |
| a · a · A · A | 2 | 0.5 | 0.5 | 0.0625 |
| a · A · a · a | 1 | 0.25 | 0.375 | 0.0625 |
| a · A · a · A | 2 | 0.5 | 0.5 | 0.0625 |
| a · A · A · a | 2 | 0.5 | 0.5 | 0.0625 |
| a · A · A · A | 3 | 0.75 | 0.375 | 0.0625 |
| A · a · a · a | 1 | 0.25 | 0.375 | 0.0625 |
| A · a · a · A | 2 | 0.5 | 0.5 | 0.0625 |
| A · a · A · a | 2 | 0.5 | 0.5 | 0.0625 |
| A · a · A · A | 3 | 0.75 | 0.375 | 0.0625 |
| A · A · a · a | 2 | 0.5 | 0.5 | 0.0625 |
| A · A · a · A | 3 | 0.75 | 0.375 | 0.0625 |
| A · A · A · a | 3 | 0.75 | 0.375 | 0.0625 |
| A · A · A · A | 4 | 1 | 0 | 0.0625 |
| Number of A copies | Number of a copies | Fraction of copies that are A | Number of equally likely copy sequences | Probability of this copy count | Probability that independent pool draws differ |
|---|---|---|---|---|---|
| 0 | 4 | 0 | 1 | 0.0625 | 0 |
| 1 | 3 | 0.25 | 4 | 0.25 | 0.375 |
| 2 | 2 | 0.5 | 6 | 0.375 | 0.5 |
| 3 | 1 | 0.75 | 4 | 0.25 | 0.375 |
| 4 | 0 | 1 | 1 | 0.0625 | 0 |
| Variation can fall while the average frequency stays the same | = |
|---|---|
| Mean A frequency across all sixteen alternatives | 0.5 |
| Mean random-pair diversity across all sixteen alternatives | 0.375 |
| Probability that only one allele remains in the pool | 0.125 |
| Probability that both alleles remain in the pool | 0.875 |
H = 2p(1 − p)
Mean frequency and retained variation answer different questions
Original idealised neutral diploid population N = 2 with 4 autosomal copies at one locus, fixed nonoverlapping generations, independent replacement copy sampling, no mutation, migration or selection. Displayed sequences are four next-generation copy positions, not DNA base sequences or measured individual genotypes. H concerns independent random draws from each resulting pool, not an observed heterozygote count.
Worked example
Keep the population and allele probability accounts distinct
For independent survival, the probability that all two members fail to survive is 0.5 × 0.5 = 0.25. For four it is 0.5⁴ = 0.0625. Their expected surviving counts are 2 × 0.5 = 1 and 4 × 0.5 = 2. A different shared-event model could give different zero probabilities even with the same marginal chance for each member.
In the separate neutral copy account, all sixteen length-four A/a sequences are equally likely. A-copy counts 0, 1, 2, 3, 4 occur in 1, 4, 6, 4, 1 sequences. Their frequencies 0, 0.25, 0.5, 0.75, 1 average to 0.5 when weighted by these probabilities. Either single-allele endpoint has probability 2/16 = 0.125.
The independent different-copy probabilities for the resulting pools are 0, 0.375, 0.5, 0.375, 0. Their weighted mean is (4 × 0.375 + 6 × 0.5 + 4 × 0.375)/16 = 0.375, below initial 0.5. This is variation lost across alternative pools, not sixteen successive generations or a measured number of heterozygous individuals. Neither probability account supplies a current species risk category.
Check your understanding
Why does an expected surviving count of one not establish that the two-member population will retain a member?
Show reasoning
The expectation averages four different outcomes with survivor counts 0, 1, 1, 2. The zero outcome has probability 0.25 under the stipulated independence conditions. An average can be positive even when complete loss is possible; a shared-event model would require a different account.
How can mean A frequency stay 0.5 while genetic variation falls in the neutral model?
Show reasoning
A-rich and a-rich outcomes balance in the mean frequency, but each shifted pool has less balanced variation, and single-allele endpoints have none. Averaging their different-copy probabilities gives 0.375 rather than 0.5. Mean allele frequency does not measure retained diversity within each outcome.
Does a small population with low variation necessarily require the same conservation action as every other small population?
Show reasoning
No. Threats, demographic trends, genetic history, habitat, potential gene-flow consequences and the decision’s ecological and social conditions differ. Census size and one genetic measure do not supply a universal safe threshold, a guaranteed fate or an appropriate practical action.
What to carry forward
Separate the event, boundary and horizon: population loss, allele loss, expected abundance and extinction-risk assessment are different accounts.
References
Resources supporting this lesson’s explanations and examples.
- Risks of Population Extinction from Demographic and Environmental Stochasticity and Random Catastrophes (The American Naturalist 142(6):911–927, 1993; doi:10.1086/285580)Russell LandeOriginal population-risk theory · PDF
- Quantification of Extinction Risk: IUCN’s System for Classifying Threatened Species (Conservation Biology 22(6):1424–1442, 2008; doi:10.1111/j.1523-1739.2008.01044.x)Georgina M. Mace and colleaguesOriginal risk-assessment analysis
- IUCN Red List of Threatened Species (Official reference to Red List categories)International Union for Conservation of NatureOfficial conservation reference
- An introduction to the mathematical structure of the Wright–Fisher model of population genetics (Theory in Biosciences 132:73–82, 2013; doi:10.1007/s12064-012-0170-3)Tat Dat Tran, Julian Hofrichter and Jürgen JostOriginal population-genetic theory
- Genetic rescue of small inbred populations: meta-analysis reveals large and consistent benefits of gene flow (Molecular Ecology 24(11):2610–2618, 2015; doi:10.1111/mec.13139)Richard FrankhamResearch abstract