A sample can contain fewer heterozygotes than Hardy–Weinberg proportions predict even when every source population fits those proportions. The missing idea is population structure: combining records does not make the organisms one randomly mating population.
We will count alleles, compare two fictional populations, pool them and identify exactly where the apparent deficit appears. Then we will change the sample sizes and test the explanation. You need fractions and a little algebra; the biological assumptions will stay beside the calculation.
Continue from The Gold Standard of Genetics, within the Gold Standard of Education: learning that transfers into life and the How X Works learning library. All numerical datasets below are invented teaching examples.
Find the explanation you need
If the pooled calculation looks convincing but the biological conclusion feels too large, begin with the worked counts. If the numbers are secure, go directly to the limits and transfer task.
Full chapter index · Practice and checking · Genetics subject guide
Chapter contents
Locate the hidden assumption · Chapters 1–2
Follow the worked counts · Chapters 3–4
Explain the deficit · Chapters 5–6
Test the biological claim · Chapters 7–8
Practise with changed data · Chapters 9–10
Questions, sources and connected reading · Chapters 11–12
CHAPTER 1 OF 12 · Locate the hidden assumption
1. Start with the population the model describes
Suppose an investigator collects fictional pond organisms at two separate sites. A spreadsheet places every organism in one table. That makes a combined sample. It does not establish that individuals from the two ponds mate randomly with each other.
Hardy–Weinberg proportions answer a biological model question: what genotype proportions follow from random union of alleles at a specified locus? Applying those proportions to a pooled frequency adds an assumption about which alleles can meet.
The Wahlund effect describes a heterozygote deficit relative to that pooled expectation when groups with different allele frequencies are combined. The comparison is with the hypothetical pooled mating model, not with the groups’ own expectations. A report that leaves out that reference point makes the result sound more mysterious than it is.
Our teaching example uses an autosomal diploid locus, two identifiable alleles and directly observed genotypes. It makes no inference about a human community or an individual’s ancestry.
Ask what merging actually did
Imagine placing a transparent sleeve over two completed genotype tables. The sleeve lets you read one total, but it cannot change the reproductive history that produced the rows underneath. This is the simplest way to keep the statistical operation separate from the biological process. Pooling is an operation on records; random union is a statement about how the next genotype combinations form.
Before opening a formula, finish this sentence: “The model treats these allele copies as able to meet because …”. If the only answer is that the files came from the same project, the biological justification is missing. An administrative boundary can contain several mating populations, and a mating population can cross a convenient administrative boundary.
The same warning applies through time. Combining two cohorts collected years apart can conceal changing allele frequencies even at one location. The example below uses two ponds to keep the source distinction visible. Its mathematical point concerns a mixture of different frequencies, rather than a special property of ponds. Retaining site and date columns makes later investigation possible; deleting them during data cleaning can remove the information needed to explain the result.
CHAPTER 2 OF 12 · Locate the hidden assumption
2. Build the allele and genotype quantities
Call the alleles A and a. AA and aa are homozygotes; Aa is a heterozygote. Each diploid individual contributes two allele copies to the count.
If a sample contains x individuals of genotype AA, y of Aa and z of aa, its size is N = x + y + z. The A frequency is p = (2x + y)/(2N). The a frequency is q = 1 − p. A frequency is not a count of organisms carrying at least one A.
Under random union with the same allele frequencies in the contributing gamete pools, expected genotype proportions are p², 2pq and q². The middle term has two routes: A then a, or a then A.
The Memorial University population-genetics notes distinguish sample, gene pool and randomly interbreeding unit. Those distinctions are the foundation of this example. Proportions in a finite observed sample need not match expectations exactly.
Check the denominator before the model
In Pond North, 96 individuals carry at least one A: the 64 AA plus 32 Aa organisms. Dividing 96 by 100 gives the carrier proportion, 0.96. Counting A copies instead gives 160 out of 200, or 0.80. Both are legitimate quantities, but substituting the first for p would answer a different question. A learner who can distinguish them has repaired an important foundation before tackling population structure.
There is another useful reverse check. If genotype proportions are 0.64, 0.32 and 0.04, recover p as 0.64 + half of 0.32 = 0.80. Each heterozygote contributes one of its two copies to A. Recover q independently as 0.04 + half of 0.32 = 0.20, then check p + q = 1.
This counting step does not require Hardy–Weinberg proportions. We can count alleles in a sample that fits the model, fails it, or has not been tested. The assumption enters when we turn those frequencies into predicted genotype proportions. Keeping counting and prediction in separate lines prevents a circular explanation in which the model is used to manufacture the observations it is supposed to compare.
For clarity, we have chosen invented counts that match each source’s model exactly. Real data generally will not be this tidy.
| Source | AA | Aa | aa | A-allele frequency |
|---|---|---|---|---|
| Pond North | 64 | 32 | 4 | 0.80 |
| Pond South | 4 | 32 | 64 | 0.20 |
| Pooled sample | 68 | 64 | 68 | 0.50 |
In Pond North, p = (2 × 64 + 32)/200 = 160/200 = 0.80. Its expected heterozygote proportion is 2 × 0.80 × 0.20 = 0.32. Among 100 individuals, that is 32 heterozygotes.
In Pond South, p = (2 × 4 + 32)/200 = 40/200 = 0.20. Its expected heterozygote proportion is also 0.32, again giving 32 individuals.
Notice what changed and what did not. The common allele switches between ponds. The heterozygote proportion stays the same because exchanging p and q does not change 2pq. Neither pond has a heterozygote deficit in this constructed comparison.
Recover the whole comparison
For North, the model predicts 0.80² = 0.64 AA and 0.20² = 0.04 aa, alongside 0.32 Aa. South reverses the two homozygote proportions. Checking all three classes matters: a heterozygote-only calculation can conceal a transcription error in a homozygote column. The counts in each row must sum to 100, and each row’s allele counts must sum to 200.
An exact match in this fictional dataset is a teaching convenience. Even an exact match in an observed dataset would not prove every condition of a long-term equilibrium model. Several processes can produce similar counts, and sampling can make different underlying situations look alike. The narrow statement is that these observations match these proportions at this locus and sampled stage.
For a quick explanation task, cover the frequency column. Ask a learner why North and South can have different common alleles yet the same heterozygote proportion. They should be able to explain the two opposite pairings, A with a and a with A. If they only say “the formula gives it,” return to the combinations before adding pooled data. The numerical symmetry becomes much easier to remember once its meaning is clear.
CHAPTER 4 OF 12 · Follow the worked counts
4. Pool the records and find the apparent deficit
The combined sample has 200 individuals and 400 allele copies. Its A count is 2 × 68 + 64 = 200, so pooled p = 0.50. If we now use a random-union model at p = 0.50, the expected heterozygote proportion is 0.50.
| Genotype | Observed among 200 | Pooled Hardy–Weinberg expectation |
|---|---|---|
| AA | 68 | 50 |
| Aa | 64 | 100 |
| aa | 68 | 50 |
Only 64 heterozygotes were observed, compared with 100 under that pooled model. The shortfall is 36 individuals, or 36/200 = 0.18 of the sample. The observed proportion is 0.32, while the pooled expectation is 0.50.
No organism changed genotype when we merged the files. The observation stayed fixed. The comparison model changed: it now permits cross-source allele combinations that the separate-source calculation did not require.
An appropriate conclusion is: “These fictional counts show a pooled heterozygote deficit consistent with the known separation of the two sources.” “The organisms must be inbred” would go beyond the example’s evidence.
Account for the excess homozygotes too
The pooled model expects 50 AA, 100 Aa and 50 aa. The observed counts exceed the two homozygote expectations by 18 each and fall short of the heterozygote expectation by 36. The differences sum to zero because the sample still contains 200 individuals. This is a redistribution relative to a comparison model, not a claim that some organisms disappeared.
A descriptive normalisation makes the reference issue especially clear. Define D = 1 − Hobs/Hpooled for this calculation. Here D = 1 − 0.32/0.50 = 0.36. That is a 36 per cent shortfall relative to the pooled heterozygote expectation. Naming this number does not tell us whether its source is subdivision, related mating, technical error or another process. The label attached to a statistic cannot supply the missing causal evidence.
Notice the distinction between an effect size and uncertainty. Our 36-individual shortfall describes the constructed sample. A statistical test would ask a further question about departures expected under a specified sampling model. It would not, merely by returning a small probability, identify the biological cause of that departure. First make the comparison meaningful; then select an inferential method appropriate to it.
CHAPTER 5 OF 12 · Explain the deficit
5. Change the sample weights without changing either source
Now sample 300 North organisms and 100 South organisms with the same source proportions. North contributes 192 AA, 96 Aa and 12 aa; South contributes 4 AA, 32 Aa and 64 aa.
The combined counts become 196 AA, 128 Aa and 76 aa. The A frequency is (392 + 128)/800 = 0.65. The observed heterozygote proportion remains 128/400 = 0.32.
The pooled model now predicts 2 × 0.65 × 0.35 = 0.455. Its apparent deficit is 0.455 − 0.32 = 0.135, smaller than the earlier 0.18.
The biological source proportions did not change. The contribution of each source to the sample changed. A comparison across surveys therefore needs the sampling weights as well as the headline heterozygosity. Otherwise a changed recruitment pattern can be mistaken for a changed biological process.
Sampling weight is not automatically population weight
The unequal example gives North three quarters of the sampled individuals. That is the correct weight for describing this sample. It is not automatically North’s share of every organism living in the two ponds. A field team might have deliberately oversampled an accessible site. Estimating a wider target population would require suitable information about that target and the recruitment design.
This distinction helps with a common reporting trap. Suppose Survey One collects equal numbers and Survey Two collects three times as many from North. An analyst can correctly report that their sample summaries differ. Claiming that the underlying source populations changed would require more evidence because the sampling mixture already explains the shift in the pooled frequency and deficit.
Keep source size, sample size and reproductive contribution in separate columns. One concerns how many organisms belong to a source, another how many entered the dataset, and a third how much that source contributes to the next generation. They coincide only under additional conditions. Our calculations deliberately use sample weights. The next-generation example later supplies its own gamete-pool assumption rather than borrowing those weights silently.
CHAPTER 6 OF 12 · Explain the deficit
6. See why averaging and multiplying do not commute
Let pᵢ be the A frequency in source i and wᵢ its sample weight, with weights summing to one. Write p̄ = Σwᵢpᵢ for the pooled frequency.
If every source follows its own Hardy–Weinberg proportions, the weighted heterozygote proportion is Hwithin = Σwᵢ[2pᵢ(1 − pᵢ)]. The pooled expectation is Hpooled = 2p̄(1 − p̄).
Subtracting gives Hpooled − Hwithin = 2[Σwᵢpᵢ² − p̄²] = 2 Varw(p). This derivation concerns the constructed model, not a significance test for finite samples.
For equal weights and frequencies 0.80 and 0.20, the weighted variance is 0.09. Twice that is 0.18, exactly our calculated deficit. For weights 0.75 and 0.25, it is 0.0675; twice that is 0.135.
You can also see the boundary case. If every source has the same p, the variance is zero. Pooling those source expectations produces no Wahlund deficit.
Follow the algebra one step at a time
Expand the within-source expression as 2Σwᵢpᵢ − 2Σwᵢpᵢ². The first sum is 2p̄. Expand the pooled expression as 2p̄ − 2p̄². Subtracting cancels the shared 2p̄ term, leaving twice the weighted mean square minus twice the square of the weighted mean. That final difference is the weighted variance of the source frequencies.
For two sources with weights w and 1 − w, the same result reduces to 2w(1 − w)(p₁ − p₂)². At a fixed frequency contrast, equal weights maximise the expression because w(1 − w) is largest at one half. If one source’s weight approaches zero, the effect approaches zero: the sample is effectively coming from one source.
This gives two independent checks without repeating the full genotype table. A contrast of 0.60 and equal weights gives 2 × 0.25 × 0.36 = 0.18. With weights 0.75 and 0.25 it gives 2 × 0.1875 × 0.36 = 0.135. The equation also shows why the sign of the contrast does not matter: the difference is squared. Relabelling North and South cannot change the biological comparison.
An unexplained heterozygote deficit is a question, not a diagnosis. Population subdivision is one candidate. Non-random mating within a source, selection between the life stages sampled, or a genotyping problem can also require investigation.
Ask whether the relevant groups were identified independently of the result. Splitting a dataset repeatedly until every subgroup looks neat is not a reliable discovery method. Biological sampling design and technical quality need their own evidence.
The Dyer Lab explanation of Wahlund effects discusses how sampling structure can produce a misleading inference of inbreeding. Our different numerical example makes that ambiguity visible without estimating an individual’s relatedness.
A finite dataset adds chance variation. Testing departures from expected proportions requires a method suited to the data and sampling design. The tidy equality inside each invented pond was chosen to expose one mechanism; it is not a requirement for real field data.
Separate two components of a total deficit
Let Hᵢ now denote the actual heterozygote proportion in source i, which need not equal 2pᵢ(1 − pᵢ). Define its within-source shortfall as dᵢ = 2pᵢ(1 − pᵢ) − Hᵢ. Straight algebra gives Hpooled − ΣwᵢHᵢ = 2 Varw(p) + Σwᵢdᵢ. This identity separates the frequency-mixture component from the weighted within-source departures.
For a fresh fictional example, retain equal weights and p values 0.80 and 0.20, but let each source have heterozygosity 0.28 instead of 0.32. Each within-source shortfall is 0.04. The pooled shortfall is now 0.50 − 0.28 = 0.22: the 0.18 mixture component plus 0.04 from the within-source departures. A single pooled number would hide that distinction.
The identity organises an explanation; it does not diagnose the 0.04 component. To investigate that component, check organism biology, mating structure, sampled stage and genotype quality. If relatives were sampled together, the observations can also have a dependence structure that a simple independent-individual analysis ignores. More rows do not necessarily supply the amount of independent information their count suggests.
CHAPTER 8 OF 12 · Test the biological claim
8. Distinguish pooling from a new generation of mating
Combining adult observations leaves their genotypes unchanged. Bringing populations into actual reproductive contact is a different event.
In a model where all contributing gametes now form one random-union pool at p = 0.50, the next zygotes have expected proportions 0.25 AA, 0.50 Aa and 0.25 aa. The earlier pooled adults still had a heterozygote proportion of 0.32.
This is why the life stage and generation belong in the question. An adult sample, a seed collection and newly formed zygotes need not describe the same stage of the process.
Do not turn this mathematical comparison into a universal forecast. Different reproductive contributions, continued separation, survival differences and other conditions can change the outcome. The useful transfer is to ask which biological event, if any, justifies replacing separate source frequencies with a single mating model.
Specify the contributing gametes
The pooled adults have p = 0.50 under equal sample weights. For the new-generation prediction to use that value, the actual gamete contributions must also give p = 0.50 in the model. If North contributes three quarters of the relevant gametes and South one quarter, their frequencies instead give p = 0.65. Equal random-union frequencies in the two gamete pools would then predict 0.4225 AA, 0.455 Aa and 0.1225 aa among the new zygotes.
There is no contradiction between those predictions and the earlier counts. They describe different biological stages under different contribution assumptions. Adult sampling weights are not a substitute for knowing how reproduction occurred. Even a well-labelled collection of adults does not, by itself, provide the genotype distribution of a future generation.
This is a useful place to practise conditional language without becoming vague. Say “If the gamete pools have these frequencies and unite randomly under this model, these are the expected proportions.” That statement is precise and testable. Saying simply “mixing restores equilibrium” discards the conditions that make the calculation valid and can imply a biological transition that nobody has observed.
CHAPTER 9 OF 12 · Practise with changed data
9. Transfer the idea to a less dramatic contrast
Try two equally represented fictional sources with A frequencies 0.60 and 0.40. Each is at its own model expectation.
The heterozygote proportion in either source is 2 × 0.60 × 0.40 = 0.48. The pooled frequency is 0.50, producing an expectation of 0.50. The deficit is therefore 0.02.
The mechanism is unchanged, but the size is much smaller. Saying “there are two sources” does not determine the magnitude; their allele-frequency contrast and sample weights matter.
Now change the context again: both sources have p = 0.60. The pooled expectation and the weighted within-source expectation are both 0.48. Different locations alone do not create this effect. The calculation needs different allele frequencies at the locus being studied.
Try a third source
Consider three equally represented fictional sources with p values 0.20, 0.50 and 0.80. Assume each follows its own random-union expectation. Their expected heterozygote proportions are 0.32, 0.50 and 0.32, so the weighted within-source value is 1.14/3 = 0.38. The pooled p is 0.50 and its heterozygote expectation is 0.50. The resulting deficit is 0.12.
Check through the variance route. The deviations from the mean are −0.30, 0 and 0.30. Their squared values average to 0.06, and doubling gives 0.12. Adding the intermediate-frequency source has reduced the deficit relative to our first two-source example, despite increasing the number of sources. Source count alone is therefore a poor shortcut for effect size.
An extreme boundary case makes the mechanism easy to see: one source is entirely AA and another entirely aa, with equal representation. Both source expectations have zero heterozygotes, while the pooled p is 0.50 and its random-union expectation is 0.50. This is a mathematical limiting case, not a claim about typical natural populations. Use it to test whether an explanation can survive when the contrast is made unmistakable.
CHAPTER 10 OF 12 · Practise with changed data
10. Check a population-genetics explanation
Before calculating, identify the locus, organism, ploidy, sampled life stage and source groups. Record how the organisms entered the dataset and whether the groups represent separate mating contexts.
Then reconstruct the allele counts from genotypes. Keep the denominator 2N visible. Calculate within-source expectations before the pooled comparison, and distinguish expected proportions from observed counts.
For an independent check, return to the unequal sample above. Without reusing its answer, find the a frequency and expected aa proportion. The results are q = 0.35 and q² = 0.1225; among 400 individuals the expectation is 49 aa. The observed aa count is 76.
Explain the discrepancy in a sentence that names the reference model. A strong answer says why the pooled comparison differs and which additional evidence would be needed to identify an unknown mechanism. Merely announcing “not in equilibrium” leaves the central reasoning unfinished.
Write an answer someone else can audit
A complete short report can contain four sentences. Identify the sample and source labels. State the observed and expected heterozygote proportions with their denominators. Explain which pooled-mating assumption creates the contrast. Finally, say which evidence would distinguish that explanation from alternatives in an unknown dataset. This order gives the reader enough information to check the reasoning rather than merely accepting a diagnostic label.
Try detecting an error in a proposed answer: “North has 96 A-bearing organisms, so p is 0.96; pooling therefore reveals strong inbreeding.” The first clause switches from allele frequency to carrier proportion. The final clause then assigns a cause without examining source structure. Correcting only the final word would leave the numerical foundation broken. Diagnose the first unstable connection and rebuild from there.
For delayed practice, use the three-source case after putting this article aside. Reconstruct its answer from source frequencies without consulting the earlier paragraph. Then explain why doubling every source sample size leaves the model proportions unchanged, although an appropriate analysis of sampling uncertainty can change. This separates the structural result from the amount of evidence available to estimate it.
CHAPTER 11 OF 12 · Questions and continuation
11. Questions that often cause confusion
Does a deficit prove inbreeding?
No. Our example creates a deficit through pooling alone. Real data require examination of structure, mating and measurement before assigning a mechanism.
Does the effect change allele frequencies when files are merged?
Pooling produces a weighted overall frequency. It does not alter the alleles or genotypes of any individual.
Must every pooled sample show this exact deficit?
No. The derivation assumes source Hardy–Weinberg proportions at a two-allele locus. Finite sampling and departures within sources add other variation.
Is a 0.18 deficit an 18 per cent reduction?
It is a difference of 0.18 in proportions, or 18 percentage points. Relative to the pooled expectation of 0.50, it is a 36 per cent shortfall. State which comparison you mean.
What should I learn first if the algebra feels difficult?
Rebuild the allele-count denominator and the two routes to Aa. Then revisit one source before pooling. The missing connection is usually more specific than being “weak at Genetics.”
Can the same sample be described at more than one scale?
Yes, provided each description retains its question. We can report the pooled allele frequency as a sample summary and report the separate source frequencies as descriptions of its composition. The error appears when the pooled summary is silently treated as evidence of one randomly mating population. Aggregation is useful; the unsupported biological assumption is what needs attention.
Do fractional expected counts mean the calculation failed?
No. In the unequal example, an expectation is a model average across possible samples, not a requirement that every observed count be an integer equal to it. An expected 182 heterozygotes happens to be whole here; other sample sizes can produce fractional expectations. Round only for an appropriate reporting purpose, retaining enough precision for the calculation.
Does a zero mixture component show that the biology is simple?
No. Equal source frequencies at one tested locus make this particular mixture component zero. That does not establish that the sources have identical frequencies everywhere, that mating is random within them, or that the observation method is error-free. A result at one locus answers a locus-specific question. State that scope rather than turning a zero in one calculation into a claim that every other explanation has been excluded.
Biological model and interpretation
Memorial University: Principles of Population Genetics supports allele counting, random-union proportions and the distinction between samples and mating units. Dyer Lab: Wahlund Effects explains the sampling-structure interpretation. Links checked on 21 October 2026. The pond counts, weighting comparison, derivation and practice tasks here are original teaching constructions.
Build the wider Genetics foundation
Use the Genetics root guide for genes, alleles and evidence, and Genetics, Probability and Inheritance Patterns for the wider mathematical toolkit.
Compare different evidence questions
Mitochondrial Bottlenecks and Sampling Variation examines a different sampling mechanism. Three-Point Mapping and Hidden Double Crossovers addresses what marker observations conceal. These models answer different biological questions; neither substitutes for population structure.
Continue into responsible interpretation
Teaching Ancient DNA Analysis connects sampled genetic evidence with population-history limits. Return to learning that transfers into life or explore How X Works. Here “gold standard” names an educational benchmark, not an official certification.
Compare a different hidden history
Population pooling changes a reference expectation; mutation timing changes descendant counts. Compare those mechanisms in The Gold Standard of Genetics | Fluctuation Tests and the Difference Between Mutants and Mutations.
Extend the evidence check
Compare the single-locus pooling question here with Linkage Disequilibrium, D′ and the Limits of Marker Prediction, which examines associations between loci. For a separate mechanism that can change observed genotype proportions after transmission, use Segregation Distortion and the Survival Filter Behind an Inheritance Ratio.