Quantitative genetics: from phenotype to breeding value

Two breeding candidates can have the same measured phenotype yet differ in the genetic merit they are expected to pass to offspring. One may have grown well because it received favourable alleles. The other may have reached the same size because it received better nutrition. Selection needs a way to separate these sources of resemblance and variation at the population level.

The gene-action section describes how alleles can act additively, through dominance within a locus, or through interactions among loci. Its environmental-effects section explains why phenotype is not determined by genotype alone. Quantitative genetics translates those biological mechanisms into statistical quantities that describe variation among individuals in a defined population.

A phenotype is a measured outcome

For individual \(i\), a useful teaching decomposition is

\[ P_i = \mu + A_i + D_i + I_i + E_i. \]

Here \(P_i\) is the measured phenotypic value and \(\mu\) is the population mean under the stated measurement conditions. The remaining terms are deviations from that mean:

  • \(A_i\) is the additive genetic value formed from population-specific average effects of the alleles carried by the individual.
  • \(D_i\) is a dominance deviation caused by the combination of two alleles at the same locus.
  • \(I_i\) represents epistatic deviations caused by interactions among loci.
  • \(E_i\) represents environmental and other non-genetic deviations included in this simplified model.

The complete genotypic value in this decomposition is \(G_i=A_i+D_i+I_i\). Dominance and epistasis can be biologically important, but their genotype combinations are not transmitted intact in the same predictable way as average allele effects. Meiosis transmits alleles, while segregation and recombination can change the within-locus and between-locus combinations in the offspring.

This equation is a model, not a claim that the five quantities can be read directly from one animal. Their definitions depend on the population, allele frequencies, environments, and measurement scale. A different decomposition might include management groups, maternal effects, genotype-by-environment interaction, or measurement error explicitly rather than placing them in \(E_i\).

Breeding value targets transmissible average effects

An individual’s breeding value is its additive genetic value \(A_i\) relative to a specified population mean. It summarises the average effects of alleles that the individual can transmit, rather than the full phenotype or complete genotypic value. This is why two individuals with the same phenotype can have different breeding values.

The word additive describes the statistical contribution of average allele effects. It does not mean that dominance or epistasis is absent from the biology. Average effects depend partly on allele frequencies and genetic background, so a breeding value is population-specific. It can change when the reference population changes, even though the individual’s DNA does not.

A true breeding value is normally latent. Breeders predict it from phenotype, pedigree, genomic markers, or a combination of these records. In the project dataset, tbv_qtl_snp is available because the data were simulated and the data-generating effects are known. Empirical breeding data do not ordinarily reveal true breeding values.

Variance describes differences among individuals

The phenotypic variance, written \(\sigma_P^2\), describes how measured phenotypes vary among individuals in the defined population and environment. The additive genetic variance, \(\sigma_A^2\), describes variation among their breeding values. Dominance and epistatic deviations can similarly contribute \(\sigma_D^2\) and \(\sigma_I^2\).

If all model components are uncorrelated, a simplified variance decomposition is

\[ \sigma_P^2 = \sigma_A^2 + \sigma_D^2 + \sigma_I^2 + \sigma_E^2. \]

That sum is not automatic. Covariance terms are required when genetic and environmental components are associated or when the statistical design does not separate them. For example, superior genotypes receiving systematically better management can create genotype-environment covariance.

The narrow-sense heritability of the trait is

\[ h^2 = \frac{\sigma_A^2}{\sigma_P^2}. \]

It is a dimensionless population parameter for a stated trait, population, environment, and time. It is not the genetic fraction of one individual’s phenotype. A value of \(h^2=0.5\) does not mean that half of an animal’s body weight was caused by genes. It means that, under the model used, additive genetic differences account for half of the phenotypic variance among individuals in that population.

Heritability can change when allele frequencies, environmental variation, management, or the sampled population changes. A high heritability also does not imply that environmental improvement is ineffective. Heritability describes variation among individuals, not whether a trait can respond to a change applied to the whole population.

The project data contain a target and a realised sample

The simulator standardised the SNP-based true breeding values to sample variance 1 and generated environmental deviations using a target heritability of 0.5. It also added a fixed sex effect to the phenotype. The following calculation deliberately compares the sample variance of the true breeding values with the sample variance of the raw phenotype.

Code
d <- readRDS("../../demo-data/main/demo_data.rds")
sigma_a2 <- var(d$pheno$tbv_qtl_snp)
sigma_p2 <- var(d$pheno$y_cont_qtl_snp)
realised_ratio <- sigma_a2 / sigma_p2
selection_differential <- 5
response_from_target <- d$meta$h2_target * selection_differential
c(
  sigma_a2 = sigma_a2,
  sigma_p2 = sigma_p2,
  realised_ratio = realised_ratio,
  target_h2 = d$meta$h2_target,
  response_from_target = response_from_target
)
            sigma_a2             sigma_p2       realised_ratio 
           1.0000000            1.7744463            0.5635561 
           target_h2 response_from_target 
           0.5000000            2.5000000

The additive sample variance is 1, the raw phenotypic sample variance is approximately 1.774, and their ratio is approximately 0.564. That ratio is not a fitted heritability estimate. The target value 0.5 determined the environmental variance used by the generator before one finite sample was drawn. The realised raw phenotype also contains the fixed sex effect, and sample covariances among genetic, environmental, and sex contributions need not equal zero. Those features prevent the raw sample ratio from reproducing the generator’s target exactly.

In empirical data, \(\sigma_A^2\) is not obtained by taking the variance of a known true breeding value column. A statistical model must estimate variance components using pedigree or genomic relationships and an explicit design for fixed and random effects. Later pages develop that model-based route.

Selection differential and expected response

Suppose the population mean for a trait is 50 units and the selected parents have mean 55. The selection differential is

\[ S = 55-50 = 5. \]

The breeder’s equation uses narrow-sense heritability to predict the change in the offspring generation mean:

\[ R=h^2S. \]

With the simulator’s target \(h^2=0.5\), the expected response is \(R=0.5(5)=2.5\) units. The predicted offspring mean would therefore be 52.5 under the assumptions of this calculation. The R chunk above reproduces the 2.5-unit response as response_from_target.

The equation is a one-generation expectation, not a guarantee for every offspring. Its simple form assumes that the parent-offspring relationship is adequately represented by additive inheritance, the heritability applies to the selection population and environment, and selection does not immediately change the relevant parameters enough to invalidate the approximation. Assortative mating, inbreeding, genotype-by-environment interaction, maternal effects, or strong changes in allele frequencies can require a richer model.

Exercises

  1. For a hypothetical individual, let \(\mu=10\), \(A_i=1.2\), \(D_i=-0.3\), \(I_i=0.1\), and \(E_i=0.5\). Calculate its genotypic and phenotypic values.

The genotypic value is \(G_i=1.2-0.3+0.1=1.0\) relative to the mean. The phenotypic value is \(P_i=10+1.0+0.5=11.5\). Only \(A_i=1.2\) is the breeding value in this decomposition.

  1. A population has \(\sigma_A^2=4\) and \(\sigma_P^2=10\). Calculate \(h^2\) and state one conclusion that the result does not support.

The narrow-sense heritability is \(h^2=4/10=0.4\). It does not mean that 40% of one individual’s phenotype is genetic. The ratio describes variance among individuals in the specified population and environment.

  1. If \(h^2=0.25\) and selected parents exceed the population mean by 3 units, what response does the breeder’s equation predict?

The selection differential is \(S=3\), so \(R=h^2S=0.25(3)=0.75\) units. This is an expected change in the population mean under the equation’s assumptions, not a prediction for each offspring.

Breeding value defines the quantity that selection seeks to improve. The next page explains how genome-wide markers and phenotypes from a training population can be used to predict that quantity for selection candidates.