Why Mate Similarity Changes Genetic Resemblance
Under random mating, the expected additive genetic contribution an offspring receives from one parent is one half of that parent’s breeding value. That fact gives the familiar parent–offspring genetic covariance of one half the additive genetic variance when the parents are unrelated and their breeding values are uncorrelated. But mates need not be genetically uncorrelated. If people choose partners partly on the basis of a heritable trait, similarity in their phenotypes can create similarity in their breeding values. The offspring then receives contributions from two parents whose breeding values tend to move together.
The key distinction is between phenotypic assortment and genetic covariance between mates. The correlation between partners’ observed traits is not automatically their genetic correlation: each phenotype also contains environmental variation. Under specific assumptions, however, the phenotypic correlation implies a predictable covariance between parental breeding values. This tutorial derives that result and follows its effect into the covariance between a parent’s breeding value and the offspring’s additive genotype value.
Model and Definitions
Consider a single quantitative trait measured on a common scale. Write an individual’s phenotype as \(P=A+E\), where \(A\) is the additive breeding value and \(E\) is the residual contribution to phenotype. Assume that both are centered at zero, \(\operatorname{Cov}(A,E)=0\), and that the population additive variance is \(V_A=\operatorname{Var}(A)\). The phenotypic variance is \(V_P=\operatorname{Var}(P)\), and narrow-sense heritability on this scale is \(h^2=V_A/V_P\). Here \(h^2\) is a variance ratio, not a statement that a particular fraction of an individual’s phenotype is genetic.
The derivation will use an additive model with no cross-mate environmental covariance and no pre-existing relatedness between mates. To connect \(r_P\) to \(r_A\), also assume a linear projection of breeding value on phenotype, \(E[A\mid P]=h^2P\), and that the residual from this projection does not covary across mates. Joint normality of \(A\) and \(P\) is one sufficient condition for the stated conditional expectation, but it is not the only possible one. These assumptions separate genetic mate resemblance induced by phenotypic assortment from shared environments, population structure, and other sources of covariance.
For transmission, assume Mendelian inheritance under the additive model. Conditional on the parents’ breeding values, the offspring’s additive value can be written as the average parental value plus a Mendelian-sampling term:
The term \(M\) represents the deviation caused by which parental alleles are transmitted. It has mean zero conditional on the parental genotypes, so it changes the offspring’s realized value but does not change the expected contribution from either parent.
From Phenotypic Assortment to Genetic Covariance Between Mates
The linear projection assumption gives a useful way to separate the predictable part of breeding value from the remaining part. Define \(U=A-h^2P\). Then \(A=h^2P+U\), and \(U\) has zero covariance with the individual’s phenotype. Under the stated cross-mate residual assumption, the terms involving \(U\) do not contribute to covariance between the partners’ breeding values. Thus:
Since \(\operatorname{Cov}(P_m,P_f)=r_PV_P\) and \(V_A=h^2V_P\), this becomes:
Dividing by \(V_A\), when \(V_A>0\), gives the implied genetic mate correlation:
This equation is conditional on the model assumptions. It says that, in this model, the genetic correlation between mates is the phenotypic correlation attenuated by heritability. If heritability is low, a given phenotypic correlation implies a smaller genetic correlation because more of the phenotypic variation is not captured by breeding value. The formula is not a universal conversion for every mating system or data set.
Worked Example: Inferring Genetic Mate Covariance
Suppose a hypothetical trait has \(V_P=100\) trait-units squared and \(V_A=40\) trait-units squared. Mates have a phenotypic correlation of \(r_P=0.50\). Assume the projection and residual conditions above, with no shared environmental covariance between mates. Find the genetic correlation and covariance between mates’ breeding values.
First calculate heritability:
The implied genetic mate correlation is:
The corresponding covariance is:
The same result follows directly from the phenotypic covariance: \(\operatorname{Cov}(P_m,P_f)=(0.50)(100)=50\), so \(h^4\operatorname{Cov}(P_m,P_f)=(0.40)^2(50)=8\) trait-units squared. The phenotypic correlation is 0.50, but the model-implied genetic correlation is 0.20, not 0.50.
Parent–Offspring Covariance Under the Two Mating Patterns
The offspring transmission equation shows exactly where mate covariance enters. Taking covariance with the first parent’s breeding value gives:
The last equality assumes both partners have the same marginal additive variance \(V_A\). The Mendelian-sampling covariance is zero by the transmission assumption. Substituting the phenotypic-assortment result yields:
The same expression holds for the second parent under symmetry. Under random mating, \(r_P=0\) and the parental breeding values are uncorrelated, so the covariance is \(V_A/2\). With positive phenotypic assortment under the stated assumptions, mate breeding values covary positively and the parent–offspring covariance exceeds \(V_A/2\). With negative assortment, it is smaller. The quantity \(A_o\) here is the offspring’s additive genotype value on the trait scale; the result is a covariance across parent–offspring pairs, not a claim that every offspring is halfway between its parents.
Worked Example: Comparing Assortative and Random Mating
For the hypothetical trait in the previous example, \(V_A=40\) trait-units squared and the implied genetic mate correlation is \(r_A=0.20\). Compute the covariance between a parent’s breeding value and the offspring’s additive genotype value under phenotypic assortment. Then compare it with random mating.
Under phenotypic assortment:
For random mating, parental breeding values are uncorrelated, so \(r_A=0\):
Thus the model predicts a parent–offspring additive covariance 4 trait-units squared larger under this assortative mating pattern. The increase is half the covariance between mates’ breeding values: \((1/2)(8)=4\). This comparison is about genetic covariance, not about a change in the offspring’s environmental conditions.
Interpreting the Formula on the Trait Scale
A common way to summarize parent–offspring resemblance is to compare the covariance with the additive variance. Under random mating, the ratio \(\operatorname{Cov}(A_m,A_o)/V_A\) is \(1/2\). Under the assortment model, it is:
This ratio describes the covariance of additive genetic values relative to \(V_A\). It is not automatically the slope from regressing an offspring’s observed phenotype on a parent’s observed phenotype. That regression also depends on phenotypic variances, environmental contributions, and the precise sampling and conditioning scheme. Keeping the genetic-value covariance distinct from a phenotypic regression prevents a change in mate pairing from being mistaken for a direct change in an individual parent’s effect.
In an additive genetic model, \(A_o\) is the trait-scale value associated with the offspring’s additive genotype. One can therefore read the result as the covariance between a parental breeding value and the offspring’s additive genotype value. It does not specify the covariance with every allele count separately, nor does it describe the distribution of multi-locus genotype combinations. Those require additional locus-specific assumptions and bookkeeping.
Worked Example: A Different Trait Scale
Consider a hypothetical trait measured in centimetres, with \(V_A=25\) square centimetres, \(h^2=0.50\), and phenotypic mate correlation \(r_P=0.40\). Assume the same additive transmission and residual conditions. Find the genetic mate covariance and the parent–offspring additive covariance.
The genetic mate correlation implied by the model is:
Therefore:
The parent–offspring covariance is:
For random mating it would be \(25/2=12.5\) square centimetres. The covariance units are squared because both variables are measured in centimetres. The genetic mate correlation, by contrast, is dimensionless.
Assumptions, Scope, and a Common Pitfall
The derivation is deliberately narrow. It assumes a single additive trait model, equal marginal breeding-value variances in the two mating groups, and a particular link between observed phenotype and mate choice. It also assumes that cross-mate environmental covariance and residual genetic covariance do not add to the covariance being attributed to phenotypic assortment. If partners share an environment that affects the trait, their phenotypes may be correlated even when their breeding values are not. Population structure or pre-existing relatedness can also produce genetic covariance between mates independently of the proposed assortment mechanism.
The results concern covariance for the specified trait scale and model. They do not, by themselves, establish how allele frequencies, genotype combinations, or genetic variances change across generations. Treating those consequences requires tracking additional features of inheritance and mating.
- Pitfall: treating phenotypic correlation as genetic correlation. The incorrect reasoning is “mates’ phenotypes correlate by 0.50, so their breeding values correlate by 0.50.” Under the stated model, the correction is \(r_A=h^2r_P\); with \(h^2=0.40\), the implied genetic correlation is 0.20.
- Pitfall: using the random-mating parent–offspring formula unchanged. The incorrect reasoning is “a parent always covaries with its offspring by \(V_A/2\).” That value follows when parental breeding values are uncorrelated. Under assortment, the additional term is \(\operatorname{Cov}(A_m,A_f)/2\).
- Pitfall: interpreting covariance as causation or as a probability. A positive parent–offspring genetic covariance describes statistical resemblance under the model; it is not the probability that a particular offspring inherits a trait, and it does not show that mating pattern alone caused an observed association.
Check Your Understanding
Answer each question using the model and assumptions stated above.
- If \(h^2=0.30\) and \(r_P=0.60\), what genetic mate correlation does the model imply?
- Why is a phenotypic mate correlation not generally equal to the genetic mate correlation?
- Starting from \(A_o=(A_m+A_f)/2+M\), identify the term that makes parent–offspring covariance differ from \(V_A/2\) under assortment.
- For \(V_A=50\) trait-units squared and \(r_A=0.10\), calculate the parent–offspring additive covariance and state its units.
- Name one environmental or population feature that could make the simple conversion \(r_A=h^2r_P\) inappropriate.