The Classical Twin Heritability Study

Monozygotic (MZ) versus Dizygotic (DZ) Twins

There are two types of twins: monozygotic (MZ, identical), and dizygotic (DZ, fraternal). The term monozygotic refers to the one-egg origin of identical twins; during early cleavage, the embryo splits into two separate cell masses each of which gives rise to a twin, both of whom originated from the same fertilized egg (zygote). Dizygotic twins arise from two distinct fertilized eggs, and hence are genetically related as any two siblings. Monozygotic twins are clones; they possess identical genotypes and DNA sequence. Dizygotic twins share 50% DNA identity-by-descent (IBD) typical for siblings.

Fig 1: Monzygotic twins originate from he same fertilized egg (zygote). Dizygotic twins originate from two separate fertilized eggs in which two distinct oocytes were fertilized by different sperm (from Figure 25.7 of "Genetics: From Genes to Genomes", 7th Edition).

For a given trait, one expects MZ twins to be more phenotypically similar than DZ twins depending on the heritability of the trait. If the trait is 100% heritable, then the MZ twins should have identical phenotypes reflecting their identical genotypes, whereas DZ twins would have a phenotypic correlation of 0.5. If the trait is 0% heritable, then the MZ twins should be as similar to each other as any two people that share a home environment such as a pair of DZ twins.

Classical Twin Equation

One can take advantage of the two types of twins to estimate heritabiilty by comparing the phenotypic similarity (covariance) between MZ and DZ twins. But let’s first start with some notation.

Previously we wrote the phenotypic value (the measured value of a trait) $z = \mu + G + E$, in which $\mu$ is the mean value of the trait in the population, $G$ is the genotypic value (contribution of genetic factors) and $E$ is the environmental value (contribution of environmental factors). $G$ and $E$ have been normalized to mean 0, and this equation is one form of the Fundamental Law of Genetics that Phenotype = Genotype + Environment.

$G$ can be broken up into an additive term, dominance term, and epistatic term. Previously we did not decompose $E$, but for twin studies we will divide it into a common environment term (environment that the two twins share) and the unique environment term (environment that they experience separately). This gives us $z = \mu + A + D + S + C + E$ in which $A$ is the additive, $D$ is the dominance, and $S$ the epistatic genotypic terms, and $C$ is the common and $E$ the unique environment terms. As a notational change we will write each of these terms as a standardized random variable with unit variance (upper-case) multiplied by a coefficient (lower-case) so that for example the additive genotypic effect previously denoted as $A$ will now be written as $aA$ with the additive variance equal to $a^2$. Finally, we will be focusing on the simplest of the twin models, the ACE model (Additive-Common-unique Environment), which ignores the dominance and epistatic terms. That leaves us with: $z = aA + cC + eE$.

Just as we calculated the phenotypic covariance between parent and offspring to estimate heritability by parent-offspring regression, we can calculate the covariance between twins ($T_1$ and $T_2$), with $\operatorname{Cov}(T_1,T_2) = \operatorname{Cov}(a_1A_1,a_2A_2) + \operatorname{Cov}(c_1C_1,c_2C_2) + \operatorname{Cov}(e_1E_1,e_2E_2) + \text{cross terms}$. For now we assume the cross-terms e.g. $\operatorname{Cov}(a_1A_1,c_2C_2)$ are 0, which we will discuss below. We also assume that the unique environment covariance is 0: $\operatorname{Cov}(e_1E_1,e_2E_2) = 0$  (e.g. the unique friends of the two twins are uncorrelated). The common environment for twins is the same so that $\operatorname{Cov}(c_1C_1,c_2C_2) = \operatorname{Var}(cC) = c^2$. This leaves us with the additive genotypic covariance which for identical twins should be $\operatorname{Cov}_{MZ}(a_1A_1,a_2A_2) = \operatorname{Var}(aA) = a^2$ because they possess the same genotype. Dizygotic twins are like any two siblings who share 50% of their genotype identity-by-descent (IBD) as do parent and offspring, and so $\operatorname{Cov}_{DZ}(a_1A_1,a_2A_2) = \frac{a^2}{2}$ using the same reasoning as when we calculated parent-offspring covariance. Thus we have: 
$ \begin{equation} \operatorname{Cov}_{MZ}(T_1,T_2) = a^2 + c^2 \text{ and }  \operatorname{Cov}_{DZ}(T_1,T_2) = \frac{a^2}{2} + c^2 \end{equation} $.
The phenotypic variance for any twin is the same as the variance for any member of the population which is $V_p = a^2 + c^2 + e^2$ given the assumptions of the ACE model. We can convert the covariances to correlations ($r$) by dividing by the variance:

$ \begin{equation} r_{MZ} = \frac{a^2 + c^2}{a^2 + c^2 + e^2} \tag{1a} \end{equation}$

$ \begin{equation} r_{DZ} = \frac{\frac{a^2}{2} + c^2}{a^2 + c^2 + e^2} \tag{1b} \end{equation}$

If we subtract the two correlations, then the common environment term cancels out, and we are left with the additive genotypic value divided by 2. Multiplying by 2 results in the following expression which is equal to narrow-sense heritability:

$\begin{equation} 2(r_{MZ} - r_{DZ}) = \frac{a^2}{a^2 + c^2 + e^2} = \frac{V_A}{V_P} = h^2 \tag{2} \end{equation} $

Thus, 2 times the difference between the monozygotic twin trait correlation and the dizygotic twin trait correlation equals narrow-sense heritability. This is the classical twin equation which is sometimes referred to as Falconer's equation.

Performing a Classical Twin Study

Operationally, performing a twin study is straightforward. One collects data from a set of MZ twins and from a set of DZ twins by measuring the trait of interest in all twins. Then one calculates the correlation between pairs of twins by plotting the trait value of twin 1 on the x-axis and the trait value of twin 2 on the y-axis followed by fitting a line to the points. The slope of the regression line $b = r \frac{\sigma_2}{\sigma_1}$ is equal to the correlation coefficient $r$ multiplied by the ratio of the standard deviation of twin 2 trait values ($\sigma_2$) over the standard deviation for twin 1 trait values ($\sigma_1$). Thus, we can obtain $r_{MZ}$ and $r_{DZ}$ by plotting the two sets of twins on separate graphs in an Excel spreadsheet. Plug the resulting values into the twin equation (Eq 2), and voila you have a heritability estimate for the trait and potentially a paper. This simplicity has resulted in twin studies becoming a cottage industry in the human genetics field.

ACE Model is an Oversimplification

But life is not so simple, or as Sasha Gusev so eloquently puts it: 

Notice all of the extra terms compared to Eqs 1a and 1b of the ACE model. There is the dominance term $D$, an epistatic term ($A*A$), a term for gene-environment interaction ($A*C$), a term that takes into account assortative mating and gene-environment correlation ($r_A$), and separate terms for the common environments of MZ versus DZ twins ($C_{MZ}$ and $C_{DZ}$). Subtracting the two correlation coefficients now results in a much more complicated expression than $\frac{a}{2}$ with extra terms that can potentially inflate the value of $2(r_{MZ} - r_{DZ})$.

It is important to emphasize that the common environment term for MZ twins ($C_{MZ}$) will not be the same as the common environment term for DZ twins ($C_{DZ}$) despite being written as the same $c$ in Eq 1. A priori, one expects the environment for MZ twins to be more similar because they are more phenotypically similar in numerous traits other than the one whose heritability is being estimated. This greater phenotypic similarity should attract and create greater environmental similarity. For example, the heritability of height is relatively large compared to other traits, and so MZ twins are more likely to be the same height, and as a result both may play basketball which can produce more common environment effects compared to DZ twins who are of different heights so that one plays basketball and the other does not. Thus, $C_{MZ}$ should be bigger than $C_{DZ}$ leading to an artifactually inflated value for $r_{MZ}$, and hence heritability. In most twin studies including  the ACE model, the assumption is $C_{MZ} = C_{DZ}$ so that it can be written just as $C$ (or $c$) which cancels out in the numerator of Eq 2.

The latter is commonly referred to as the Equal Environments Assumption (EEA), which posits that monozygotic (identical) twins and dizygotic (fraternal) twins share their common environments to the same extent. If monozygotic twins are treated more similarly or share more environmental influences than dizygotic twins, this could lead to an overestimation of heritability.

A second key assumption of the classical twin study is completely ignoring gene-environment interaction and correlation (some of the cross-terms in covariance calculation). If there is gene-environment interaction (in which the effect of genes on a trait depends on the environment, or vice versa; the $A*C$ term), or gene-environment correlation (in which genetic factors influence exposure to environmental factors), this could also upwardly bias heritability estimates. For example, if gene-environment correlations exist, then more similar genotypes can lead to more similar environments even for the "unique" environmental factors.

Finally, it is possible that non-equal environments and gene-environment interactions can feed back upon each to further increase the difference between $r_{MZ}$ and $r_{DZ}$ resulting in even more upward bias. One DZ twin ends up on the basketball team whereas the other ends up on the chess team, and this difference reinforces non-equal environments because of the different circles of friends. 

TL;DR: The classical twin study is straightforward to describe and perform but depends on numerous questionable assumptions that upwardly bias the heritability estimates, including EEA and neglecting to consider gene-environment effects.

The fundamental problem is that $a$, $c$, and $e$ in Eq 1 are latent variables that cannot be experimentally measured directly and so instead are inferred indirectly from a model (e.g. ACE model), but the model possesses suspect assumptions so that the inferred values are also suspect.

In contrast, one may adopt the more direct approach of predicting phenotype from genotype, with the accuracy of this prediction being limited by the heritability of the trait. The more heritable the trait, the more accurate the prediction. 

Given that we know the complete genotype of an individual from the DNA sequence, then we should be able to make a phenotypic prediction for any trait as long as we have collected enough data to construct a powerful enough (linear) model relating genotype to phenotype. This idea forms the basis of molecular heritability estimates which is the subject of a future post and whose values tend to be significantly lower than those from classical twin studies.

Heritability

In a previous post I described “The Fundamental Law of Genetics,” which states that phenotype (observed trait) = genotype (genetic factors) + environment (environmental factors). The quantitative version of the law is $V_P = V_G + V_E$ in which the phenotypic variance in a population equals genotypic variance (phenotypic variation caused by differences in genotype)  plus environmental variance (phenotypic variance caused by differences in environment). Then I cautioned that the equation is a simplification that omits genotype-by-environment interactions.

In a more recent post I described an alternative version of this law as $z$ (phenotypic value) $= G$ (gentoypic value) $+ E$ (residual or environmental value). A phenotypic value is a measurement of a trait like your height, and the genotypic value is the portion attributed to genotype with the remainder being attributed to environment. This value equation can be converted into the variance equation by taking the variance of both sides and then throwing out the covariance term between G and E.

Intuitively, any trait (such as blood pressure) depends on genetic factors (alleles) that you inherit from your parents as well as environmental factors (such as diet and exercise). But how much is genetics and how much is environment? The concept of heritability attempts to estimate the fraction of a trait that is due to genetics (your genotype).

More quantitatively, heritability is the  fraction of total phenotypic variance ($V_P$) that can be explained by genotypic variance ($V_G$):

$ \begin{equation} H^2 = \frac{V_G}{V_P} = \frac{V_G}{V_G + V_E} \tag{1} \end{equation} $

where $H^2$ refers to broad-sense heritability. My students asked whether they needed to take the square root to obtain $H$, but no, $H^2$ not $H$ is defined to be heritability (good question as always).

Estimating heritability is easier said than done, rife with assumptions and caveats. Further, there is a wide gap between the technical definition of heritability, and the common-sense understanding among the general public (e.g. the apple doesn't fall far from the tree) who tend to overestimate heredity and underestimate rearing. Many of these issues are beyond the scope of this post, and here I will focus on one simple approach to estimating heritability by comparing the phenotypes of parent and child.

In the early 1900s, Francis Galton gathered data highlighting the similarities between parents and children. I have reproduced Figure 8.2 from Hartl and Clark which shows a scatter plot of the weight of male offspring on the y-axis with the weight of their father on the x-axis. The fact that this trait is inherited to a certain degree is reflected in the linear relationship between the parental trait value and the offspring trait value, so that the data points can be fit with a line (i.e. linear regression, which Galton helped to invent). More importantly, this line has a positive slope so that higher offspring values correspond to higher parental values.

Approximate reproduction of Figure 8.2 from Hartl and Clark (2007). Mean weight of male offspring of flour beetle (y-axis) plotted against weight of father (x-axis).  

On the one hand, if the heritability was very low for the trait, then the data points would be randomly scattered and the best line to fit the points would be horizontal of slope 0. On the other hand, if the heritability was close to 1, then the children would have the same value of the trait as their parent (in the simplest case), and the points would lie along a line of slope 1. 

The converse is not true; points lying along a line of slope 1 do not imply that the heritability is 1. There could be environmental components which are identical for both parents and children, e.g. both grew up in wealthy households. Environmental covariance refers to related environments (e.g. for parents and offspring) leading to related phenotypes. However, if we assume that the environmental covariance is zero, we can calculate heritability from the linear regression slope.

We can simplify matters by defining the "midparent" which is the average phenotype of the Mother and Father, and plot the offspring phenotypic values (y-axis) against the midparent phenotypic values (x-axis). From statistics, the resulting regression line that best fits these points will have slope $b = \frac{\operatorname{Cov}(x,y)}{\operatorname{Var}(x)}$.

The phenotypic covariance between child ($y$) and midparent ($x$) is a measure of trait resemblance, i.e. how correlated are the two phenotypes (correlation is scaled covariance). As stated above, one can write the phenotypic value $z$ as the sum of the genotypic value $G$ (contribution of genetic factors) and the environmental value $E$ (contribution of environmental factors): $z = G + E$.

In the general case for any two individuals $x$ and $y$, the phenotypic covariance can written as: $\operatorname{Cov}(x,y)  = \operatorname{Cov}(G_x + E_x, G_y + E_y) = \operatorname{Cov} (G_x,G_y) + \operatorname{Cov}(G_x,E_y) + \operatorname{Cov}(E_x,G_y) + \operatorname{Cov}(E_x,E_y)$ where $G_x$ represents the genotypic values and $E_x$ the environmental values for individual $x$. We will assume that the genotype-environment covariances ($\operatorname{Cov}(G,E)$ terms) are 0 so that we can focus on the remaining two terms.

The term $\operatorname{Cov}(E_x,E_y)$ describes the relatedness of the environments of the two. If $x$ and $y$ were random individuals, then we would expect this covariance to be 0, but as mentioned above, for parent and child many environmental conditions are likely to be similar (“I’m raising my kid just like I was raised”) so that the covariance would not be zero for many traits. Assuming this covariance is 0 is the biggest weakness of calculating heritability from parent-offspring regression, but for now we will assume that it is 0 to continue with the blog post using the derivations outlined in Coop (2020), Hartl and Clark (2007), and Lynch and Walsh (1998).

The remaining term is the covariance of genotypic values $\operatorname{Cov} (G_x,G_y)$. As discussed in the previous post on dominance and epistasis, each genotypic value can be broken up into additive, dominance and epistatic terms: $G_x = G_{xA} + G_{xD} + G_{xS}$. We can then write $\operatorname{Cov} (G_x,G_y) = \operatorname{Cov}(G_{xA},G_{yA}) + Cov(G_{xD},G_{yD}) + \operatorname{Cov}(G_{xS},G_{yS}) +$ cross terms that are 0. For parent and offspring, $\operatorname{Cov}(G_{xD},G_{yD}) = 0$, and most of the epistasis terms are also 0 or presumed to be small (see Table 7.2 in Chapter 7, Lynch and Walsh for details). This leaves us with $\operatorname{Cov} (G_x,G_y) \sim \operatorname{Cov}(G_{xA},G_{yA})$. As a reminder, the additive term is from the linear model of summing the contributions of individual alleles that influence a trait, whereas dominance and epistasis represent nonadditive genetic contributions.

In close relatives, one can calculate the covariance of the additive genotypic values by following the descent of alleles to the respective individuals. Given a locus, an offspring inherits one allele from one parent and one allele from the other parent. The additive covariance between parent and child is 0 for the allele not inherited from the parent (correlation = 0), and is $\frac{V_A}{2}$ for the inherited allele, in which $V_A$ is the additive genotypic variance for the locus with the $\frac{1}{2}$ arising from that a locus consists of two alleles. Thus, for the locus we have $\operatorname{Cov}(G_{xA},G_{yA}) = 0 + \frac{V_A}{2} = \frac{V_A}{2}$. Similar reasoning applies to calculating the additive genotypic covariance between offspring and midparent.

The above is an example of the concept of identity-by-descent (IBD) which refers to shared, completely identical DNA segments between two or more individuals inherited from a common ancestor. A parent shares 50% of DNA with a child through IBD, and so 50% of the offspring's alleles are identical to the parent's through direct descent.

As described above, we can write the slope $b$ of the midparent-offspring regression as the covariance divided by the variance of the midparent which is $Var(x) = \frac{V_P}{2}$ because the midparent is the phenotypic average of the two parents. This leads to the following equation:

$ \begin{equation} b = \frac{\operatorname{Cov}(x,y)}{\operatorname{Var}(x)} \sim \frac{\operatorname{Cov}(G_{xA},G_{yA})}{\operatorname{Var}(x)} = \frac{V_A}{V_P} = h^2 \tag{2} \end{equation} $

The expression $h^2 = \frac{V_A}{V_P}$ is referred to as narrow-sense heritability which consists only of the additive component of the genotypic variance (omitting the dominance and epistatic terms) divided by the total phenotypic variance, and hence it is always less than or equal to broad-sense heritability. However, both theory and empirical data indicate that $h^2$ is close to $H^2$ (Benjamin, et al. 2024). 

Thus, we can calculate the narrow-sense heritability by performing linear regression of offspring phenotypic values against parental phenotypic values for a trait. The main limitation is the assumption that the environmental covariance of parent and offspring is 0 ($\operatorname{Cov}(E_x,E_y) = 0$). If the covariance is non-zero, then the estimate of $h^2$ will be upwardly biased because there is an additional positive term in the numerator of Eq 2 that shouldn't be there. Because of this suspect assumption, researchers in the past have turned to a different method using twins to estimate heritability, the subject of a future post.

TL;DR: There is broad-sense and narrow-sense heritability which describe the fraction of a trait that is attributed to genetic factors. The latter can be estimated by parent-offspring regression, but with the key assumption that environmental covariance is 0. However, it is very unlikely that the parental environment is uncorrelated with the offspring environment. The apple and the tree grow on the same patch of land.