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.

Can the Federal Government Print Money to Finance Its Debt?

Editor's note: This post (except TL;DR) was written by ChatGPT in response to my question whether the Federal government could print money to finance its debt.

Legal and Practical Limitations

• Separation of Powers: The U.S. Treasury is responsible for issuing government debt (e.g., Treasury bonds, bills, and notes) to finance deficits. It does not have the authority to print money to pay for this debt. The Federal Reserve, the central bank, manages the money supply and monetary policy and is an independent entity that does not directly finance government spending.
• Prohibition on Monetizing Debt: "Monetizing debt" refers to a central bank directly buying government bonds to fund spending. In the U.S., this is avoided to prevent inflation and ensure trust in the currency. Instead, the government borrows money by selling bonds to private investors, institutions, or foreign governments.

Federal Reserve's Role

While the Federal Reserve cannot directly finance government spending, it can indirectly influence the process:

• It purchases Treasury securities on the open market (not directly from the Treasury) as part of its monetary policy. This action injects liquidity into the economy and indirectly supports government borrowing.
• During crises like the COVID-19 pandemic, the Federal Reserve purchased large amounts of government debt to stabilize markets and support economic recovery.

Risks of Printing Money to Finance Debt

If the federal government were to print money to pay off its debt directly, it would bypass the Federal Reserve, leading to:

• Hyperinflation: Excessive money supply can devalue the currency and cause prices to rise rapidly (e.g., Zimbabwe, Venezuela).
• Loss of Confidence: Global investors might lose faith in the U.S. dollar as a stable reserve currency.
• Economic Instability: The dollar's value could plummet, and borrowing costs could spike.

Conclusion

While the federal government indirectly benefits from the Federal Reserve's monetary policy actions, it cannot and does not print money to directly finance its debt. Instead, it relies on borrowing through bond issuance, supported by a robust and independent monetary system to manage inflation and economic stability.

TL;DR: Although the Federal government cannot directly print money, it can indirectly print money via the Federal Reserve purchasing large amounts of government debt (not directly from Treasury) which technically can be rolled over (new debt purchased after maturity of old debt) indefinitely.