The common marginal trait variance is
$$
V_Y=V_A+V_C+V_E.
$$
For monozygotic pairs, shared genes and shared environment give
$$
\operatorname{Cov}(Y_{i1},Y_{i2})=V_A+V_C,
\qquad
r_{\rm MZ}=\frac{V_A+V_C}{V_Y}.
$$
For dizygotic pairs, the genetic covariance is halved while the common environment is unchanged:
$$
\operatorname{Cov}(Y_{i1},Y_{i2})=\frac12V_A+V_C,
\qquad
r_{\rm DZ}=\frac{V_A/2+V_C}{V_Y}.
$$
Subtracting cancels the common-environment variance and proves <Falconer's formula>
$$
2(r_{\rm MZ}-r_{\rm DZ})
=\frac{V_A}{V_Y}=h^2,
$$
the <heritability> under the <ACE model>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 221 / 2 / ii / Solution

{title2=$h^2$}
{wiki}

Heritability is the fraction of phenotypic variance in a specified population and environment attributed to genetic variation. It is a population parameter, not the genetic fraction of one individual's trait.


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