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Predict a population response without turning a ratio into destiny
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Full chapter index · Independent practice · Genetics subject guide
Heritability describes variation in a specified population and environment. It does not divide an individual plant into a genetic percentage and an environmental percentage. To use it well, a learner must identify which variance is being measured, distinguish broad-sense from narrow-sense heritability, and compare a predicted response to selection with an appropriately controlled next generation.
This guide uses invented plant-height examples to derive R = h²S, test how an environmental change alters the conclusion, and explain why a high heritability estimate can coexist with a large response to better growing conditions. The calculations are teaching models, not breeding recommendations or claims about human potential.
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Full chapter index
Understand the question · Chapters 1–2
Build and calculate · Chapters 3–4
Change the assumptions · Chapters 5–6
Check the evidence · Chapters 7–8
CHAPTER 1 OF 12 · Understand the question
1. Start with differences, not ingredients of one individual
Suppose a group of plants has different heights. Some differences may be associated with inherited genetic variation, while others reflect light, water, soil, developmental variation or measurement. Heritability asks about the variance of those heights within a defined population under the conditions studied.
This is different from asking why a particular plant reached 50 centimetres. A variance ratio does not allocate 30 centimetres to genes and 20 centimetres to the environment. Development depends on interacting biological processes; the statistical partition describes differences across observations under a model.
Define the population carefully. A sample of one variety in one glasshouse is not interchangeable with a collection of diverse varieties grown across several climates. The range of genetic and environmental variation differs, so the variance components and their ratio can differ. Neither estimate is automatically wrong; they may answer different questions.
Define the phenotype as well. Height at a specified age, final height, mean height across replicate plants and growth rate are different measured traits. Their units, errors and sources of variation may differ. A report that gives only “heritability of height” omits information necessary for interpretation.
For the examples below, imagine individual final plant heights measured in centimetres in a defined breeding population. All variance components are in cm². Keeping the observational unit and the squared units visible helps prevent confusion when the calculation later switches from variances to a selection differential measured in centimetres.
CHAPTER 2 OF 12 · Understand the question
2. Broad-sense and narrow-sense heritability answer different questions
In a simplified model, phenotypic variance Vp is partitioned into additive genetic variance Va, dominance variance Vd, interaction variance Vi and environmental or residual variance Ve. Under the stated assumptions, Vp = Va + Vd + Vi + Ve. More detailed models may include additional covariance terms or genotype-by-environment structure.
Broad-sense heritability is H² = Vg/Vp, where Vg includes all genetic variance represented by the model. In this simplified partition, Vg = Va + Vd + Vi. Narrow-sense heritability is h² = Va/Vp. The distinction matters because the simple breeder’s equation for response to selection with sexual reproduction uses the additive component.
“Additive” does not mean that genes operate without molecular interaction. It is a statistical description of average allele effects in a population. The contribution of a locus to additive variance depends on allele frequencies and genetic background. Biological complexity can therefore contribute to a population-level additive predictor without becoming a literal collection of independent molecular switches.
Consider the invented values Va = 30, Vd = 20, Vi = 10 and Ve = 40 cm². Then Vp = 100 cm². Broad-sense H² = 60/100 = 0.60, while narrow-sense h² = 30/100 = 0.30. Substituting 0.60 for 0.30 in the same simple response prediction would double the expected change without justification.
Neither ratio tells us which specific alleles are present in an individual or which environment would maximise growth. They summarise a model fitted to a specified population. For clonal propagation, inbred lines or selection on family means, the relevant prediction may involve a different genetic unit and variance definition. Always match the statistic to the biological design.
CHAPTER 3 OF 12 · Build and calculate
3. Selection differential and response have different baselines
Let the original population mean height be μ = 50 cm. The plants chosen as parents have a mean height μs = 60 cm. The selection differential is S = μs − μ = 10 cm. It describes how far the selected parental group lies above the original population mean on the measured phenotype.
The response R describes the change in mean phenotype in the next generation attributable to the selection comparison under the relevant design. In a simplified common-environment expectation, R = h²S. With h² = 0.30 and S = 10 cm, the predicted response is 3 cm, giving an expected next-generation mean of 53 cm on the original environmental baseline.
The selected parents averaged 60 cm, yet the predicted offspring mean is 53 cm. This does not mean that heredity has failed or that offspring lose exactly seven centimetres of parental height. The selected parental phenotype included additive inherited differences plus other components that are not transmitted as the same average breeding value in the next generation.
The baseline is crucial. Subtracting 60 from the offspring mean answers a different question from subtracting the original population mean. Likewise, comparing offspring grown in a new environment with old parents confounds inherited response with environmental change. An unselected control generation grown alongside the selected line is one way to make that distinction more visible.
The equation also needs a clear definition of selection. The simple example treats the selection differential as representative of the parental contribution under its assumed mating design. If only one sex or one parental group is selected, or selected individuals contribute unequal numbers of offspring, the calculation needs the corresponding reproductive weights and genetic model.
Weight parents by what they contribute
An arithmetic mean of selected individuals can be misleading if their reproductive contributions differ. Imagine two selected parents with heights 55 and 65 cm. Their simple mean is 60 cm. If the first contributes nine times as many gametes to the relevant selected parental pool as the second, the contribution-weighted phenotypic mean is (9 × 55 + 1 × 65)/10 = 56 cm.
Relative to an original mean of 50 cm, that weighted differential is 6 cm rather than 10 cm. Under the same simplified response assumptions with h² = 0.30, applying the weighted differential gives 1.8 cm rather than 3 cm. The example isolates weighting; a complete mating design would also specify the contributions from the other parental pool.
This calculation does not establish that taller selected parents necessarily have proportionally higher breeding values. It shows that the selection differential must describe the population that actually contributes to reproduction. Recording who was chosen and recording who contributed are distinct tasks. Unequal contribution can alter the expected response even before environmental complications are introduced.
CHAPTER 4 OF 12 · Build and calculate
4. Work through the variance and response calculation
Use the invented population with Vp = 100 cm² and Va = 30 cm². First compute h² = 30/100 = 0.30. Then compute S = 60 − 50 = 10 cm. Finally R = 0.30 × 10 = 3 cm. Each step has a different role, and keeping them separate makes an arithmetic or interpretation error easier to find.
The phenotypic standard deviation is √100 = 10 cm. The selected mean is therefore one phenotypic standard deviation above the population mean. In this specific example the standardised selection differential is 1. The additive genetic standard deviation is √30, approximately 5.477 cm, which is not the same as h² or the phenotypic standard deviation.
If the selected mean were 55 cm instead, S would be 5 cm and the simple predicted response would be 1.5 cm. If the selected mean were 45 cm, S would be −5 cm and the predicted response would be −1.5 cm. A negative response in this design represents selection toward shorter plants; it is not an invalid negative heritability.
Check the units. Va and Vp are both variances in cm², so their ratio is dimensionless. S and R are differences in means measured in centimetres. Multiplying h² by a variance or dividing the selected mean itself by h² does not produce the requested response.
The table below compares broad-sense and narrow-sense calculations in one original example. Its numbers are exact inputs to a teaching model. Real variance components are estimated with uncertainty and may depend on the fitted statistical structure. Reporting several decimal places in R would not make an uncertain h² estimate more precise.
Compare equal heritability with different scales
Consider a second invented population with Va = 3 and Vp = 10 cm². Its h² is also 0.30. The original example had Va = 30 and Vp = 100. The equal ratio does not mean the two populations have the same absolute variation or the same response under every selection rule.
If selection in each population produces the same S = 10 cm, the simple equation predicts the same R = 3 cm. But achieving that differential is a different task in the two distributions. The second population has phenotypic standard deviation √10 ≈ 3.162 cm, so 10 cm is over three standard deviations, compared with one standard deviation in the first population.
If instead the selection rule gives the same standardised differential of one phenotypic standard deviation, then S is 10 cm in the first population and about 3.162 cm in the second. Predicted responses are 3 cm and about 0.949 cm. Equal h² can therefore coexist with different absolute responses under a standardised selection comparison.
Keep the selection rule beside the heritability estimate. A statement about expected gain needs both. A high ratio with very little absolute variation may offer less scope for change than a lower ratio combined with a much larger available selection differential.
| Quantity | Value | Meaning |
|---|---|---|
| Va | 30 cm² | Additive variance |
| Vd + Vi | 30 cm² | Other genetic variance |
| Ve | 40 cm² | Environmental/residual variance |
| Vp | 100 cm² | Total variance |
| H² / h² | 0.60 / 0.30 | Different variance ratios |
| S / R | 10 cm / 3 cm | Differential / predicted response |
CHAPTER 5 OF 12 · Change the assumptions
5. Why changing the environment can change the ratio
Imagine two populations with the same additive variance Va = 30 cm² and no other genetic variance in this simplified comparison. In Environment A, environmental variance is 70 cm², so Vp = 100 and h² = 0.30. In Environment B, environmental variance is 20 cm², so Vp = 50 and h² = 0.60.
The ratio doubles even though the additive variance has not changed. A more uniform environment reduces one component of the denominator. This illustrates why a high heritability can occur when environmental differences within the observed setting are restricted. It does not establish that environmental conditions are biologically unimportant.
Means and variances are separate. Suppose improved watering raises every plant’s height by 8 cm. In that idealised uniform-shift example, the mean increases by 8 cm while the variance is unchanged. A heritability ratio could remain the same even though the environment has a large effect on every plant.
Conversely, a change in environmental variability could alter h² without changing the mean. The ratio alone therefore cannot tell whether the plants grew taller, whether the growing conditions improved, or whether a specific intervention would help. Those are questions about means, mechanisms and comparisons, not just variance fractions.
When comparing studies, write down what changed in the numerator and denominator. A statement that “heritability increased” is incomplete without the populations, environments, phenotype definitions and estimation methods. The useful interpretation may be that the study became more environmentally uniform, that genetic diversity increased, that measurement became less noisy, or that several components changed together.
CHAPTER 6 OF 12 · Change the assumptions
6. Use a control to separate selection from a new season
Return to the original mean of 50 cm and selected-parent mean of 60 cm. The simple genetic prediction is R = 3 cm. Now suppose the next generation is grown in a season that raises all heights by 8 cm relative to the original setting. An unselected control population would be expected to average 58 cm, while the selected line would average 61 cm under the example’s assumptions.
The observed difference from the old mean is 61 − 50 = 11 cm. Dividing 11 by S = 10 would give a naive realised estimate of 1.1. That calculation has mixed a seasonal effect with the selection response. It does not overturn the variance-ratio interpretation or demonstrate that the trait is more than completely inherited.
The contemporaneous comparison is 61 − 58 = 3 cm. Dividing that controlled response by S gives 3/10 = 0.30, matching the teaching input. The control does useful work because it experiences the new season without the same selection treatment.
Real controls require care. The selected and unselected lines should be comparable in ancestry, growing conditions, management and measurement. Drift, unequal sample sizes, maternal effects or a genotype-specific response to the season can complicate the difference. A control reduces a particular confounding problem; it does not remove every possible source of uncertainty.
The general lesson is to identify the counterfactual: what would the next-generation mean have been without the selection programme under the same conditions? A historical mean may be a poor substitute for that comparison. Asking this question is useful far beyond breeding because it separates an intervention’s effect from background change.
CHAPTER 7 OF 12 · Check the evidence
7. Genotype-by-environment interaction changes the transfer question
Suppose genotype A averages 40 cm in a dry environment and 60 cm in a wet environment. Genotype B averages 50 cm when dry and 55 cm when wet. In the dry setting, B is taller; in the wet setting, A is taller. The ranking reverses.
This is an invented example of genotype-by-environment interaction: the effect of environment depends on genotype, and the genotype comparison depends on environment. It is different from a uniform +8 cm shift applied to every plant. A single environmental correction cannot align both genotypes in the crossover example.
A heritability estimate from one environment does not automatically predict the response in another. Selection in the dry setting may favour genotypes that are not tallest in the wet setting. Predicting transfer requires evidence about the relationship between genetic effects across environments, not just the magnitude of h² in one of them.
Write the intended target explicitly. Is the aim improved height in a particular field, stable performance across several fields, or performance under rare stress? These aims can favour different selection criteria. A high estimate for one tightly controlled setting does not answer every target question.
For students, draw a two-column table rather than using the phrase “genes and environment interact” as a stopping point. Put the four means in the table, compare the genotypes within each environment, and compare environments within each genotype. The numerical reversal makes the limitation concrete and provides a changed-context check on an otherwise abstract definition.
| Genotype | Dry mean (cm) | Wet mean (cm) |
|---|---|---|
| A | 40 | 60 |
| B | 50 | 55 |
CHAPTER 8 OF 12 · Check the evidence
8. What an estimate requires and what family resemblance can hide
Heritability is estimated through a design, not read directly from a list of heights. Relatedness patterns, controlled crosses, replicated lines and genomic information can each contribute to an estimation strategy. The assumptions determine how observed resemblance is attributed to genetic and other sources.
For example, related plants may be grown in the same tray. If the tray receives more light, family resemblance can partly reflect a shared environment. A design that confounds family with tray cannot cleanly identify the source of the difference. Randomisation and replication across blocks help make the components distinguishable.
Similarly, averaging several measurements can reduce some measurement noise. The heritability of a replicated entry mean may be higher than that of an individual observation. Those estimates refer to different observational units. They should not be compared as though a plant’s biological inheritance changed simply because the statistic used an average.
Unequal replication, missing plots and spatial correlations complicate familiar formulas. Piepho and Möhring’s research on unbalanced plant breeding trials illustrates why a simple variance ratio is not a universally adequate predictor of selection response. The brief lesson here is to match the estimate to the actual design rather than transplanting a formula from a balanced classroom example.
An uncertainty interval is useful because h² is an estimate. Small samples, limited relatedness information and model assumptions can make it imprecise. An estimated component outside its usual theoretical range may signal sampling or fitting issues rather than a new biological law. Inspect the method and uncertainty before assigning a strong interpretation to one reported number.
Separate measurement precision from biological change
Suppose an individual measurement has additive variance 20 units² and independent measurement-error variance 20 units², with no other components in this stripped-down example. The individual-observation heritability is 20/(20 + 20) = 0.50. If the same biological value can be measured four times with independent errors and then averaged, the error variance of that mean becomes 20/4 = 5 units².
The ratio for the averaged measurement becomes 20/(20 + 5) = 0.80. The additive variance has not increased. The recording procedure has made the selection information less noisy. This is a statement about a particular error model and repeated measurement of the same underlying value, not a promise that every kind of biological repeat behaves independently.
If all four measurements share the same calibration bias, averaging will not remove that shared component. Likewise, measuring a growing plant on four different days may capture genuine developmental change rather than repeated errors around one stable value. The definition of the repeated quantity is part of the model.
This example gives students a concrete reason to ask whether a reported estimate applies to individuals, repeated-measure means or replicated genotype means. A comparison between 0.50 and 0.80 can otherwise be misread as evidence that one population is more genetically determined when the main difference is observational precision.
CHAPTER 9 OF 12 · Practise and continue
9. Why one generation does not determine the next ten
The breeder’s equation in this guide is a short-term population prediction under stated conditions. Repeated selection can change allele frequencies, additive variance, association among loci and the environment experienced by offspring. The h² and S used in the first generation need not remain constant.
Suppose a learner predicts a 3 cm gain in the first generation and then adds 3 cm in every later generation. That arithmetic is a scenario with constant response, not evidence that the population will follow it. A longer-term prediction needs an account of changing variance, selection intensity, reproductive structure and other biological constraints.
Finite populations add sampling variation. The chosen parents and the alleles transmitted to offspring are not infinitely large averages. An observed line can move above or below the expected response, and parallel selected lines may differ. Replication helps distinguish a repeatable response from an unusual trajectory.
Selection can also affect correlated traits. Taller plants might differ in flowering time, lodging risk or resource use. A one-trait response equation does not describe all consequences of a breeding decision. A student should be able to calculate the height prediction while recognising the other outcomes that remain unmeasured.
For learning, use a sequence of increasingly realistic questions: calculate one generation with fixed inputs; change the environment; introduce uncertainty in h²; then explain why repeated response may change. This preserves the clarity of the simple model while showing precisely where additional information becomes necessary.
Read a response comparison as evidence
Suppose two replicate selected lines show controlled responses of 2.5 and 3.5 cm when the model predicts 3 cm. Averaging those observed responses gives 3 cm, but the individual deviations still contain information about finite sampling and experimental variation. Reporting only the average would hide that variability; reporting only one line would overstate the certainty of its realised result.
If both selected and control lines are measured by an instrument that adds 2 cm to every reading, their same-generation difference is unchanged. If one line is measured with that instrument and the other with a correctly calibrated one, the comparison is biased. Consistent measurement therefore matters even in a design with a control.
These checks are useful when reviewing a study summary: identify the unit of replication, the contemporaneous comparison, and the measurement procedure. A response equation explains an expected mean under assumptions. A well-designed experiment supplies evidence about whether those assumptions and expectations work in the population actually studied.
CHAPTER 10 OF 12 · Practise and continue
10. Independent practice and worked explanations
Practice A: Va = 18 cm², Vd = 12 cm² and Ve = 30 cm², with no interaction variance in the stated model. The population mean is 40 cm and selected-parent mean is 48 cm. Find H², h², S and the simple expected R. Check: Vp = 60; H² = 30/60 = 0.50; h² = 18/60 = 0.30; S = 8 cm; R = 2.4 cm.
Practice B: In A, replace Ve = 30 with Ve = 90 while leaving the genetic variances unchanged. Check: Vp = 120, H² = 0.25 and h² = 0.15. If S remains 8 cm, R becomes 1.2 cm. This is a stipulated comparison; changing the environmental variance does not guarantee that actual selected parents would still have the same S.
Practice C: A selected line averages 74 units and an unselected contemporary control averages 70 units. The original population mean was 60 and S was 10. Find the controlled response and the naive historical response. Check: controlled R = 4, giving R/S = 0.40; the historical difference is 14, giving 1.4 if incorrectly treated as pure response.
Practice D: Does H² = 0.80 mean improved watering cannot substantially change the population mean? Check: no. The ratio concerns variance within the studied population and environment. A uniform environmental shift can change the mean while leaving the variance and ratio unchanged.
Practice E: A fitted h² estimate has an uncertainty range of 0.20 to 0.50 and S = 6 cm. Under the same simple model, what responses correspond to those endpoints? Check: 1.2 to 3.0 cm. These are propagated endpoint calculations, not a complete prediction interval for the realised next generation, which also contains other sources of uncertainty.
CHAPTER 11 OF 12 · Practise and continue
11. Questions that reveal the common misunderstandings
Is heritability a property of a trait everywhere?
It is a statistic for a specified population, environment, phenotype definition and estimation model. Calling a trait highly heritable without those conditions can hide substantial differences between studies. It is more informative to describe the study setting than to attach a universal percentage to the trait.
Does high heritability mean a trait cannot change?
No. A variance ratio does not measure the size of every possible environmental effect. The uniform-shift example shows how the mean can change greatly while the ratio remains stable. To assess an intervention, study the intervention and its relevant comparison directly.
Why does this response prediction use narrow-sense h²?
The simple response prediction depends on the additive component associated with average transmissible effects in the stated population and mating design. Broad-sense genetic variance contains additional components that do not enter this particular simple prediction in the same way. Other propagation and selection designs need their own definitions.
Can h² tell us how tall one offspring will be?
It supports a population-level prediction under a model, not an exact individual forecast. Offspring differ through segregation, environments, development and measurement. The expected mean of many offspring and the outcome of one plant are different prediction targets.
Does a poor realised response prove the estimate was false?
It calls for investigation. Sampling, environmental change, selection weights, genotype-by-environment interaction and model mismatch may all matter. Compare the observed response with its uncertainty and the actual design before deciding which assumption failed.
CHAPTER 12 OF 12 · Practise and continue
12. Sources and a practical interpretation checklist
For the distinction between genetic and phenotypic variation, see the US National Library of Medicine’s What is heritability? Its explanation is especially useful for checking the difference between a population statistic and an individual’s development. The plant datasets and calculations in this guide are original teaching examples.
For primary research on the design dependence of prediction, see Piepho and Möhring’s Computing Heritability and Selection Response From Unbalanced Plant Breeding Trials. For theoretical limits of simple response prediction, see Heywood’s An exact form of the breeder’s equation for the evolution of a quantitative trait under natural selection.
Before interpreting an estimate, record the population, environment, trait, age or measurement time, observational unit, genetic variance definition, mating design, uncertainty and intended prediction target. Before calculating response, define the original mean, selected-parent mean and appropriate offspring comparison. These details turn an isolated fraction into an assessable explanation.
Continue from the Genetics subject guide. Read the Wahlund effect to see why population definition matters in another genetic calculation, and linkage disequilibrium to separate association, prediction and causal interpretation.
A strong learner can finish with one carefully qualified statement: under this population, environment and selection design, these variance components predict this expected change in the mean. That statement is both more useful and more accurate than turning a heritability percentage into a fixed verdict about an individual.
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