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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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By the end, you can

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.

01

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.

02

A mean is not a persistence guarantee

Demographic stochasticity is variation arising from individual survival and reproductive events in a finite population. In a small account, a few different individual outcomes can change the whole total substantially. The independent-survival example makes that idea visible without fitting a real population model. Every member has the same survival probability 0.5, but a zero-survivor outcome is possible.

For two starting members there are four equally likely fate assignments: neither, first only, second only or both survive. Their survivor counts 0, 1, 1, 2 have mean 1. For four starting members there are sixteen assignments and mean 2; the all-loss assignment has probability 1/16 instead of 1/4. These are one-interval local accounts with no reproduction, arrival or departure, not multi-year population viability estimates.

Environmental variation and major disturbances can affect many members together, so independent events are often an incomplete model. Larger populations can remain vulnerable to shared habitat losses or sustained negative demographic contributions. Persistence estimates therefore depend on events, dependence, time horizon and uncertainty as well as size. A universal “safe number” would hide these conditions and is not supplied by this exercise.

Equal individual chances give different probabilities of losing every member

Equal individual chances give different probabilities of losing every memberIndependent survival with probability one half yields a complete set of alternative outcomes for two-member and four-member populations over one interval.○ Survives the interval; × Does not survive the intervalAssigned starting population of 2 membersInitialmember1Initialmember2Survivingmembers10213142Alternative complete fate assignmentShared initial member identity within this panelProbability of this survivor countSurviving members: 0Surviving members: 1Surviving members: 200.250.50.751Mean surviving count across alternatives: 1Probability of no surviving members: 0.25Assigned starting population of 4 membersInitialmember1Initialmember2Initialmember3Initialmember4Survivingmembers102131425162728391102112123132143153164Alternative complete fate assignmentShared initial member identity within this panelProbability of this survivor countSurviving members: 0Surviving members: 1Surviving members: 2Surviving members: 3Surviving members: 400.250.50.751Mean surviving count across alternatives: 2Probability of no surviving members: 0.0625Survives the interval: 0.5Individual survival events are stipulated independentLocal loss and global species extinction have different boundaries
A mean of one survivor is not a guarantee that someone survives. Shared environmental events would violate this independence account, and larger size does not ensure persistence.
Read the diagram in words

For N = 2 the four alternatives give survivor-count probabilities 0:1/4, 1:2/4, 2:1/4 and mean 1. For N = 4 the sixteen alternatives give 0:1/16, 1:4/16, 2:6/16, 3:4/16, 4:1/16 and mean 2. Every member still has survival probability 0.5; the probability of zero differs as 0.25 versus 0.0625.

○ Survives the interval; × Does not survive the interval

Assigned starting population of 2 members
Alternative complete fate assignmentInitial member 1Initial member 2Surviving membersP(○/×)
1Does not survive the intervalDoes not survive the interval00.25
2Does not survive the intervalSurvives the interval10.25
3Survives the intervalDoes not survive the interval10.25
4Survives the intervalSurvives the interval20.25
Probability of this survivor count
Surviving membersNumber of equally likely fate assignmentsProbability of this survivor count
010.25
120.5
210.25
Assigned starting population of 2 members
Assigned starting population of 2 members=
Assigned starting population of 2 members2
Mean surviving count across alternatives1
Probability of no surviving members0.25
Assigned starting population of 4 members
Alternative complete fate assignmentInitial member 1Initial member 2Initial member 3Initial member 4Surviving membersP(○/×)
1Does not survive the intervalDoes not survive the intervalDoes not survive the intervalDoes not survive the interval00.0625
2Does not survive the intervalDoes not survive the intervalDoes not survive the intervalSurvives the interval10.0625
3Does not survive the intervalDoes not survive the intervalSurvives the intervalDoes not survive the interval10.0625
4Does not survive the intervalDoes not survive the intervalSurvives the intervalSurvives the interval20.0625
5Does not survive the intervalSurvives the intervalDoes not survive the intervalDoes not survive the interval10.0625
6Does not survive the intervalSurvives the intervalDoes not survive the intervalSurvives the interval20.0625
7Does not survive the intervalSurvives the intervalSurvives the intervalDoes not survive the interval20.0625
8Does not survive the intervalSurvives the intervalSurvives the intervalSurvives the interval30.0625
9Survives the intervalDoes not survive the intervalDoes not survive the intervalDoes not survive the interval10.0625
10Survives the intervalDoes not survive the intervalDoes not survive the intervalSurvives the interval20.0625
11Survives the intervalDoes not survive the intervalSurvives the intervalDoes not survive the interval20.0625
12Survives the intervalDoes not survive the intervalSurvives the intervalSurvives the interval30.0625
13Survives the intervalSurvives the intervalDoes not survive the intervalDoes not survive the interval20.0625
14Survives the intervalSurvives the intervalDoes not survive the intervalSurvives the interval30.0625
15Survives the intervalSurvives the intervalSurvives the intervalDoes not survive the interval30.0625
16Survives the intervalSurvives the intervalSurvives the intervalSurvives the interval40.0625
Probability of this survivor count
Surviving membersNumber of equally likely fate assignmentsProbability of this survivor count
010.0625
140.25
260.375
340.25
410.0625
Assigned starting population of 4 members
Assigned starting population of 4 members=
Assigned starting population of 4 members4
Mean surviving count across alternatives2
Probability of no surviving members0.0625

Survives the interval: 0.5

Individual survival events are stipulated independent

Local loss and global species extinction have different boundaries

Original closed populations with initial N = 2 or N = 4, independent identical survival probability 0.5 and no births, immigration or emigration over one defined interval. All binary outcomes are equally likely within a panel. Zero survivors means local loss in this included boundary, not global species extinction; no later recolonisation or viability threshold is supplied.

03

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

Variation can fall while the average frequency stays the sameFour independent replacement draws from a half-A, half-a pool yield all sixteen equally likely copy sequences in one assigned generation.Initial four-copy poolAAaaFraction of copies that are A: 0.5; Probability that independentpool draws differ: 0.5Next-generation copy positions in the supplied orderNumberof AcopiesH1aaaa002aaaA10.3753aaAa10.3754aaAA20.55aAaa10.3756aAaA20.57aAAa20.58aAAA30.3759Aaaa10.37510AaaA20.511AaAa20.512AaAA30.37513AAaa20.514AAaA30.37515AAAa30.37516AAAA40H: Probability that independent pool draws differProbability of this copy countNumber of A copies:0; Number of equallylikely copysequences: 1Number of A copies:1; Number of equallylikely copysequences: 4Number of A copies:2; Number of equallylikely copysequences: 6Number of A copies:3; Number of equallylikely copysequences: 4Number of A copies:4; Number of equallylikely copysequences: 100.250.50.751Mean A frequency across all sixteen alternatives: 0.5Mean random-pair diversity across all sixteen alternatives: 0.375Probability that only one allele remains in the pool: 0.125Probability that both alleles remain in the pool: 0.875H = 2p(1 − p)Mean frequency and retained variation answer differentquestions
The sixteen sequences are alternatives, not a sixteen-generation trajectory. A neutral allele can be lost without being harmful, and an allele loss does not by itself establish population loss.
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=
Initial four-copy poolA · A · a · a
N2
2N4
Fraction of copies that are A0.5
Probability that independent pool draws differ0.5
Next-generation copy positions in the supplied order
Next-generation copy positions in the supplied orderNumber of A copiesFraction of copies that are AProbability that independent pool draws differP(A/a)
a · a · a · a0000.0625
a · a · a · A10.250.3750.0625
a · a · A · a10.250.3750.0625
a · a · A · A20.50.50.0625
a · A · a · a10.250.3750.0625
a · A · a · A20.50.50.0625
a · A · A · a20.50.50.0625
a · A · A · A30.750.3750.0625
A · a · a · a10.250.3750.0625
A · a · a · A20.50.50.0625
A · a · A · a20.50.50.0625
A · a · A · A30.750.3750.0625
A · A · a · a20.50.50.0625
A · A · a · A30.750.3750.0625
A · A · A · a30.750.3750.0625
A · A · A · A4100.0625
Probability of this copy count
Number of A copiesNumber of a copiesFraction of copies that are ANumber of equally likely copy sequencesProbability of this copy countProbability that independent pool draws differ
04010.06250
130.2540.250.375
220.560.3750.5
310.7540.250.375
40110.06250
Variation can fall while the average frequency stays the same
Variation can fall while the average frequency stays the same=
Mean A frequency across all sixteen alternatives0.5
Mean random-pair diversity across all sixteen alternatives0.375
Probability that only one allele remains in the pool0.125
Probability that both alleles remain in the pool0.875

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

  1. 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.

  2. 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.

  3. 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.

  1. 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 · PDFms.mcmaster.caRead source
  2. 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 analysisconbio.onlinelibrary.wiley.comRead source
  3. IUCN Red List of Threatened Species (Official reference to Red List categories)International Union for Conservation of NatureOfficial conservation referenceiucn.orgRead source
  4. 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 theorypmc.ncbi.nlm.nih.govRead source
  5. 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 abstractpubmed.ncbi.nlm.nih.govRead source