chapter-04.tex

\section{The phenotypic resemblance between relatives}

We can use our understanding of the sharing of alleles between relatives
to understand the phenotypic resemblance between relatives in
quantitative phenotypes. We can then use this to understand the
evolutionary change in quantitative phenotypes in response to selection. \\

\subsection{A simple additive model of a trait}
Let's imagine that the genetic component of the variation in our trait
is controlled by $L$ autosomal loci that act in an additive manner. The frequency of allele $1$ at locus $l$ is $p_l$, with each copy of
allele $1$ at this locus increasing your trait value by $a_l$ above
the population mean.
The phenotype of an individual, let's call her $i$, is $X_i$.
Her genotype at SNP $l$, is
$G_{i,l}$. Here $G_{i,l}=0,~1,$ or $2$  represents the number of copies of allele $1$ she
has at this SNP. Her expected phenotype, given her genotype, is then
\begin{equation}
\E (X_i | G_{i,1},\cdots,G_{i,L}) =\mu + X_{A,i} = \mu+\sum_{l=1}^L G_{i,l} a_{l} \label{pheno_geno}
\end{equation}
where $\mu$ is the mean phenotype in our population, and $X_{A,i}$ is
the deviation away from the mean phenotype due to her genotype. Now in reality the genetic phenotype is a function of the
expression of those alleles in a particular environment. Therefore, we
can think of this expected phenotype as being an average across a set
of environments that occur in the population. \\

%\gc{NEED to resolve $\mu$ in above equation}


When we measure our individual's observed phenotype we see
\begin{equation}
X_i =   \mu+X_{A,i} + X_{E,i} \label{pheno_geno_environ}
\end{equation}
where $X_E$ is the deviation from the mean phenotype due to the
environment. This $X_E$ included the systematic effects of the environment
our individual finds herself in and all of the noise during
development, growth, and the various random insults that life throws
at our individual. If a reasonable number of loci contribute to
variation in our trait then we can approximate the distribution of
$X_{A,i}$ by a normal distribution due to the central limit theorem (see Figure \ref{fig:QT1}). Thus if we can
approximate the distribution of the effect of environmental variation
on our trait ($X_{E,i}$) also by a normal distribution, which is
reasonable as there are many small environmental effects, then the
distribution of phenotypes within the population ($X_i$) will be
normally distributed (see Figure \ref{fig:QT1}).\\

\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{figures/QT1.png}
\end{center}
\caption{The convergence of the phenotypic distribution to a normal
  distribution. Each of the three histograms shows the distribution of
the phenotype in a large sample, for increasing large numbers of loci ($L$). I have simulated each individual's
phenotype following equation \ref{pheno_geno} and \ref{pheno_geno_environ}. Specifically I simulate each
individuals biallelic genotype at $L$ loci, assuming Hardy-Weinberg proportions
and that the allele is at 50\% frequency. I assume that all of the
alleles have equal effects and combine them additively together. I then
add an environmental contribution, which is normally distributed with
variance $0.05$. Note that in the left two pictures you can see peaks
corresponding to different genotypes.} \label{fig:QT1}
\end{figure}

Note that as this is an additive model we can decompose eqn. \ref{pheno_geno_environ} into the
effects of the two alleles at each locus, in particular we can rewrite
it as
\begin{equation}
X_i = \mu + X_{iM}+X_{iP} +X_{iE}
\end{equation}
where $X_{iM}$ and $X_{iP}$ are the contribution to the phenotype of
the allele that our individual received from her mother (maternal
alleles) and father (paternal alleles) respectively. This will come in
handy in just a moment when we start thinking about the phenotype covariance of relatives.\\

Now obviously this model seems silly at first sight as alleles don't
only act in an additive manner, as they interact with alleles at the
same loci (dominance) and at different loci (epistasis). Later we'll
relax this assumption, 
however, we'll find that if we are interested in evolutionary change
over short time-scales it is actually only the ``additive
component'' of genetic variation that will (usually) concern us. 
We will define this more formally later on, but for the moment 
we can offer the intuition that parents only get to pass on a single
allele at each locus on to the next generation. As such, it is the
effect of these transmitted alleles, averaged over possible matings,
that is an individual's average contribution  to the next generation
(i.e. the additive effect of the alleles that their genotype consists of).



\subsection{Additive genetic variance and heritability}
As we are talking about an additive genetic model we'll talk about the
additive genetic variance ($V_A$), the variance due to the additive
effects of segregating genetic variation. This is a subset of the total genetic
variance if we allow for non-additive effects. \\

The variance of our phenotype across individuals ($V$) can write this as
\begin{equation}
V = Var(X_A) + Var(X_E) = V_A+V_E
\end{equation}
in writing this we are assuming that there is no covariance between $X_{G,i}$
and $X_{E,i}$ i.e. there is no covariance between genotype and
environment. \\

Our additive genetic variance can be written as
\begin{equation}
V_A = \sum_{l=1}^L Var(G_{i,l} a_{l})
\end{equation}
where $Var(G_{i,l} a_{l})$ is the contribution to the additive
variance among individuals of the $l$ locus. Assuming random mating we
can write our additive genetic variance as
\begin{equation}
V_A = \sum_{l=1}^L a_{l}^2 2 p_l(1-p_l)
\end{equation}
where the $ 2 p_l(1-p_l)$ term follows the binomial sampling of two
alleles per individual at each locus. \\

\paragraph{The narrow sense heritability}
We would like a way to think about what proportion of the variation
in our phenotype across individuals is due to genetic differences as
opposed to environmental differences. Such a quantity will be key in
helping us think about the evolution of phenotypes. For example, if
variation in our phenotype had no genetic basis then no matter how
much selection changes the mean phenotype within a generation
the trait will not change over generations. \\

We'll call the proportion of the variance that is genetic the
heritability, and denote it by $h^2$. We can then write this as
\begin{equation}
h^2 = \frac{Var(X_A)}{V} = \frac{V_A}{V}
\end{equation}
remember that we thinking about a trait where all of the alleles act
in a perfectly additive manner. In this case our heritability $h^2$ is
referred to as the narrow sense heritability, the proportion of the
variance explained by the additive effect of our loci.
When we allow dominance
and epistasis into our model we'll also have to define the broad sense
heritability (the total proportion of the phenotypic variance
attributable to genetic variation).\\

The narrow sense heritability of a trait is a useful quantity, indeed
we'll see shortly that it is exactly what we need to understand the
evolutionary response to selection on a quantitative phenotype. We can
calculate the narrow sense heritability by using the resemblance between
relatives. For example, if our phenotype was totally environmental we
should not expect relatives to resemble each other any more than random
individuals drawn from the population. Now the obvious caveat here is
that relatives also share an environment, so may resemble each other
due to shared environmental effects. \\

\subsection{The covariance between relatives}
So we'll go ahead and calculate the covariance in phenotype between two individuals
($1$ and $2$) who have a phenotype $X_1$ and $X_2$ respectively.
\begin{equation}
Cov(X_1,X_2) =
Cov\left((X_{1M}+X_{1P}+X_{1E}),((X_{2M}+X_{2P}+X_{2E}) \right)
\end{equation}
We can expand this out in terms of the covariance between the various
components in these sums.\\

To make our task easier we (and most analyses) will assume two things
\begin{enumerate}
\item that we can ignore the covariance of the environments
between individuals (i.e. $Cov(X_{1E},X_{2E})=0$)
\item that we can ignore the covariance
between the environment variation experience by an individual and the
genetic variation in another individual (i.e. $Cov(X_{1E},(X_{2M}+X_{2P}))=0$).
\end{enumerate}

The failure of these assumptions
to hold can severely undermine our estimates of heritability, but we'll
return to that later. Moving forward with these assumptions, we can
write our phenotypic covariance between our pair of individuals as
\begin{equation}
Cov(X_1,X_2) =
Cov((X_{1M},X_{2M})+Cov(X_{1M},X_{2P})+Cov(X_{1P},X_{2M})
+Cov(X_{1P},X_{2P}) \label{cov_rels_1}
\end{equation}
This is saying that under our simple additive model we can see the
covariance in phenotypes between individuals as the covariance between
the allelic effects in our individuals. We can use our results about
the sharing of alleles between relatives to obtain these terms.
But before we write down the general case lets quickly work through some
examples. \\


\paragraph{The covariance between Identical Twins}
Lets first consider the case of a pair of identical twins from two
unrelated parents. Our pair of twins share their maternal and paternal
allele identical by descent ($X_{1M}=X_{2M}$ and $X_{1P}=X_{2P}$). As their maternal and
paternal alleles are not correlated draws from the population,
i.e. have no probability of being $IBD$ as we've said the parents are unrelated, the
covariance between their effects on the phenotype is zero  
(i.e. $Cov(X_{1P},X_{2M})=Cov(X_{1M},X_{2P})=0$). In that case
eqn. \ref{cov_rels_1} is
\begin{equation}
Cov(X_1,X_2) = Cov((X_{1M},X_{2M})+Cov(X_{1P},X_{2P}) = 2Var(X_{1M})
= V_A
\end{equation}
Now in general identical twins are not going to be super helpful for
us in estimating $h^2$ as under models with non-additive effects
identical twins have higher covariance than we'd expect as they
resemble each other also because of the dominance effects as they
don't just share alleles they share their entire genotype.\\

\paragraph{The covariance in phenotype between mother and child}.
If the mother and father are unrelated individuals (i.e. are two
random draws from the population) then the mother and a child share
one allele IBD at each locus (i.e. $r_1=1$ and $r_0=r_2=0$). Half the
time our mother transmits her paternal allele to the child, in which
case $X_{P1}=X_{M2}$ and so $Cov(X_{P1},X_{M2})=Var(X_{P1})$ and all
the other covariances in eqn. \ref{cov_rels_1} zero, and half
the time she transmits her maternal allele to the child
$Cov(X_{M1},X_{M2})=Var(X_{M1})$ and all the other terms zero. By this
argument $Cov(X_1,X_2) = \half Var(X_{M1}) + \half Var(X_{P1}) = \half
V_A$. \\

\paragraph{The covariance between general pairs of relatives under an
additive model}
The two examples make clear that to understand the covariance between
phenotypes of relatives we simply need to think about the alleles they
share IBD. Consider a pair of relatives ($1$ and $1$) with a probability $r_0$,
$r_1$, and $r_2$ of sharing zero, one, or two alleles IBD
respectively. When they share zero alleles
$Cov((X_{1M}+X_{1P}),(X_{2M}+X_{2P}))=0$, when they share one allele
$Cov((X_{1M}+X_{1P}),(X_{2M}+X_{2P}))=
Var(X_{1M})=\frac{1}{2}V_A$, and when they share two alleles $Cov((X_{1M}+X_{1P}),(X_{2M}+X_{2P}))=
V_A$. Therefore, the general covariance between two
relatives is
\begin{equation}
Cov(X_1,X_2) = r_0 \times 0 + r_1 \frac{1}{2}V_A + r_2  V_A =
2 F_{1,2} V_A  \label{additive_covar_general_rellys}
\end{equation}\\
So under a simple additive model of the genetic basis of a phenotype
to measure the narrow sense heritability we need to measure the
covariance between a set of pairs of relatives (assuming that we can remove the effect of
shared environmental noise). From the covariance between relatives we
can calculate $V_A$, we can then divide this by the total phenotypic
variance to get $h^2$. \\
%One way potentially to get somewhat around the
%shared environmental effect is to use paternal half-sibs as they share a

\begin{tcolorbox}
\begin{question}
{\bf A)} In polygynous blackbird populations (i.e. males mate with
several females), paternal half-sibs can be identified.  Suppose that
the covariance of tarsus lengths among half-sibs is 0.25 $cm^2$ and
that the total phenotypic variance is 4 $cm^2$.  Use these data to
estimate $h^2$ for tarsus length in this population. \\

{\bf B)} Why might paternal half-sibs be preferable for measuring
heritability than maternal half-sibs? 
\end{question}
\end{tcolorbox}

Another way that we can estimate the narrow sense heritability is
through the regression of child's phenotype on the parental mid-point
phenotype. The parental mid-point phenotype is simple the average of
the mum and dad's phenotype. Denoting the child's phenotype by $X_{kid}$ and mid-point
phenotype by $X_{mid}$ so that if we take the regression $X_{kid} \sim X_{mid}$ this
regression has slope $\beta = Cov(X_{kid},X_{mid})/Var(X_{mid})$.
The covariance of $Cov(X_{kid},X_{mid})=\half
V_A$, and $Var(X_{mid}) = \half V$ as by taking the average of the
parents we have halved the variance, such that the slope of the
regression is
\begin{equation}
\beta_{mid,kid}= \frac{Cov(X_{kid},X_{mid})}{Var(X_{mid})} = \frac{V_A}{V} = h^2
\end{equation}
i.e. the regression of the child's phenotype on the parental midpoint
phenotype is an estimate of the narrow sense heritability. This is a
common way to estimate heritability, although it doesn't bypass the
need to control for environmental correlations between relatives. \\

Our regression allows us to attempt to predict the phenotype of the
child given the phenotypes of the parents; how well we can do this depends on the
slope. If the slope is close to zero then the parental phenotypes hold no
information about the phenotype of the child, while if the slope is
close to one then the parental mid-point is a good guess at the child's
phenotype.\\
\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{figures/QT2.pdf}
\end{center}
\caption{Regression of parental mid-point phenotype on child's
  phenotype. The three panels show decreasing levels of environmental
  variance ($V_E$) holding the additive genetic variance constant ($V_A=1$). 
 In these figures I simulate $100$ loci, as described in
 the caption of Figure \ref{fig:QT1}. I simulate the genotypes and
 phenotypes of the two parents, and then simulate the child's genotype
following mendelian transmission. The blue line shows $x=y$ the red
line shows the best fitting linear regression line. }
\end{figure}

More formally the expected phenotype of the child given the parental
phenotypes is
\begin{equation}
\E(X_{kid} | X_{mum},X_{dad}) = \mu +
\beta_{mid,kid}(X_{mid} - \mu) =\mu + h^2(X_{mid} - \mu)  \label{predict_kid}
\end{equation}
this follows from the definition of linear regression. So to find the
child's predicted phenotype we simply take the mean phenotype and add
on the difference between our parental mid-point multiplied by our
narrow sense heritability. \\


\begin{question}
Briefly explain Galton’s observation of the regression towards
mediocrity in light of Mendelian inheritance. 
\end{question}

\paragraph{Estimating additive genetic variance across a variety of
  different relationships.}

In many natural populations we may have access to individuals of a
range of different relationships to each other (through monitoring
of the paternity of individuals), but relatively few individuals of a
given relationship (e.g. sibs). We can try and use this information as
fully as possible in a mixed model framework. Considering equation
\ref{pheno_geno_environ} we can write an individual's phenotype $X_i$
 as 
\begin{equation}
X_i =  \mu  + X_{A,i} + e_i 
\end{equation}
where $e_i \sim N(0,V_E)$ and $X_{A,i}$ is normally distributed across
individuals with covariance matrix $V_A A$ where the the entries for
a pair of individuals i and j are 
$A_{ij}= 2 F_{i,j}$ and $A_{ii}= 1$. Given the matrix $A$ we can estimate $V_A$. We can
also add fixed effects into this model to account for generation
effects, additional mixed effects could also be included to account
for shared environments between particular individuals (e.g. a shared nest).
This is sometimes called the ``animal model'', and also goes by the
name of variance components analysis.


\subsection{The response to selection}
Evolution by natural selection requires:
\begin{enumerate}
\item Variation in a phenotype
\item That survival is non-random with respect to this phenotypic
variation.
\item That this variation is heritable.
\end{enumerate}
Points 1 and 2 encapsulate our idea of Natural Selection, but evolution by natural
selection will only occur if the 3rd condition is met. It is the
heritable nature of variation that couples change within a generation
due to natural selection, to change across generations (evolutionary
change). \\

Lets start by thinking about the change within a generation due
to directional selection, where selection acts to change the mean
phenotype within a generation. For example, a decrease in mean height within a
generation, due to taller organisms having a lower chance of surviving
to reproduction than shorter organisms. Specifically, we'll denote our mean phenotype at
reproduction by $\mu_S$, i.e. after selection has acted, and our mean
phenotype before selection acts by $\mu_{BS}$. This second quantity may be hard to
measure, as obviously selection acts throughout the life-cycle, so it
might be easier to think of this as the mean phenotype if selection
hadn't acted. So the mean phenotype changes within a generation is $\mu_{S} - \mu_{BS}= S$.  \\

We are interested in predicting the distribution of phenotypes in next
generation, in particular we are interested in the mean phenotype in
the next generation to understand how directional selection has
contributed to evolutionary change. We'll denote the mean phenotype in
offspring, i.e. the mean phenotype in the next generation before selection acts,
as $\mu_{NG}$. The change across generations we'll call the response
to selection $R$ and put this equal to $\mu_{NG}- \mu_{BS}$. \\

The mean phenotype in the next generation is
\begin{equation}
\mu_{NG} = \E \left( \E(X_{kid} | X_{mum},X_{dad}) \right)
\end{equation}
where the outer expectation is over the randomly mating of individuals
who survive to reproduce. We can use eqn. \ref{predict_kid} to obtain
an expression for this
\begin{equation}
\mu_{NG} = \mu_{BS} +
\beta_{mid,kid} ( \E(X_{mid}) - \mu_{BS})
\end{equation}
so to obtain $\mu_{NG}$ we need to compute $\E(X_{mid})$ the expected
mid-point phenotype of pairs of individuals who survive to
reproduce. Well this is just the expected phenotype in the individuals
who survived to reproduce ($\mu_{S}$), so
\begin{equation}
\mu_{NG} = \mu_{BS} +
h^2 (\mu_S - \mu_{BS})
\end{equation}
So we can write our response to selection as
\begin{equation}
R = \mu_{NG} -\mu_{BS}  =
h^2 (\mu_S - \mu_{BS}) = h^2 S \label{breeders_eqn}
\end{equation}
So our response to selection is proportional to our selection
differential, and the constant of proportionality is the narrow sense
heritability. This equation is sometimes termed the Breeders
equation. It is a statement that the evolutionary change across
generations ($R$) is proportional to the change caused by directional selection
within a generation, and the strength of this relationship is
determined by the narrow sense heritability. \\

Using the fact that $h^2=V_A/V$ we can rewrite this in a different form as
\begin{equation}
R= V_A \frac{S}{V}
\end{equation}
i.e. our response to selection is the additive genetic variance of our
trait ($V_A$) multiplied by the change within a generation as a
fraction of the total phenotypic variance ($S/V$, sometimes called the
the selection gradient $\beta$).\\

If our selection pressure is sustained over many generations we can
use our breeders equation to predict the response. If we are willing
to assume that our heritability does not change and we maintain a constant selection
gradient, then after $n$ generations our phenotype mean will have
shifted 
\begin{equation}
n h^2 S
\end{equation}
i.e. our population will keep up a linear response to selection.

\begin{figure}
\begin{center}
\includegraphics[width=0.8\textwidth]{figures/QT3.png}
\end{center}
\end{figure}

A change in mean phenotype within a generation occurs because of the
differential fitness of our organisms. To think more carefully about this change within a
generation lets think about a simple fitness model where our phenotype affects the
viability of our organisms (i.e. the probability they survive to
reproduce). The probability that an individual has a phenotype $X$
before selection is $p(X)$, so that the mean phenotype before
selection is
\begin{equation}
\mu_{BS} = \E[X] =  \int_{-\infty}^{\infty} x p(x) dx
\end{equation}
The probability that an organism with a phenotype $X$ survives to
reproduce is $w(X)$, and we'll think about this as the fitness of
our organism. The probability distribution of phenotypes in those who
do survive to reproduce is
\begin{equation}
\P(X | \textrm{survive}) =  \frac{p(x) w(x)}{
\int_{-\infty}^{\infty} p(x) w(x) dx}.
\end{equation}
where the denominator is a normalization constant which ensures that
our phenotypic distribution integrates to one. The denominator also
has the interpretation of being the mean fitness of the population,
which we'll call $\wbar$, i.e.  
\begin{equation}
\wbar =  
\int_{-\infty}^{\infty} p(x) w(x) dx.
\end{equation}

Therefore, we can write the mean phenotype in those who survive to
reproduce as
\begin{equation}
\mu_S = \frac{1}{\wbar}\int_{-\infty}^{\infty} x p(x) w(x) dx
\end{equation}

If we mean center our population, i.e. set the phenotype before
selection to zero, then
\begin{equation}
S= \frac{1}{\wbar}\int_{-\infty}^{\infty} x p(x) w(x) dx
\end{equation}
if $\mu_S=0$.  Inspecting this more closely we can see that $S$ has
the form of a covariance between our phenotype $X$ and our fitness
$w(X)$ ($Cov(X,w(X))$). Thus our change in mean phenotype is directly a measure of the
covariance of our phenotype and our fitness. Rewriting our breeder's
equation using this observation we see
\begin{equation}
R = \frac{V_A}{V}  Cov(X,w(X))
\end{equation}
we see that the response to selection is due to the fact that our
fitness (viability) of our organisms/parents covaries with our phenotype, and
that our child's phenotype is correlated with the parent phenotype.

\begin{question}
A student has been studying barnacle populations and has found the mean height (before selection) to be 10 mm and the variance in height to be 4 mm2.  After a stormy season (one generation), the mean of the population in the next generation before selection has increased to 12 mm.  In a breeding experiment, the student found the slope of the regression between parental midpoint and child to be 0.2.  Assuming the observed change was caused by selection, what was the mean height of the parents that survived the stormy season?  
\end{question}

\begin{question}
A population of red deer were trapped on Jersey (an island off of
England) during the last inter-glacial period. From the fossil record
we can see that the population rapidly adapted to their new
surroundings, presumably due to reduced predation and limited
food. Within 6,000 years they evolved from an estimated mean weight of
the population of 200kg to an estimated mean weight of 36kg (a 6 fold
reduction! True story see: \href{http://www.nature.com/nature/journal/v342/n6249/abs/342539a0.html}{Lister, A.M. Nature 1989}). Using a current
day population on the mainland you estimate that the generation time
of red deer is 5 years and that the narrow sense heritability of the
phenotype is 0.5. (Assuming discrete generations).\\
{\bf A)}	Estimate the mean change per generation in the mean body weight. \\

{\bf B)}	Estimate the change in mean body weight caused by selection within a generation.\\

{\bf C)}	What do you have to assume to perform the calculations in B. Assuming we only have fossils from the founding population and the population after 6000 years, should we assume that the calculations accurately reflect what actually occurred within our population?
\end{question}

\subsection{Multiple traits.}

Traits often covary with each other, due to both environmentally
induced effects (e.g. due to the effects of diet on multiple traits)
and due to the expression of underlying genetic covariance between
traits. In turn this genetic covariance can reflect pleiotropy, a
mechanistic effect of an allele on multiple traits (e.g. variants that
effect skin pigmentation often effect hair color) or the genetic
linkage of loci independently affecting multiple traits. If we are
interested in evolution over short time-scales we can (often) ignore
the genetic basis of this correlation. 

Consider two traits $X_{1,i}$ and $X_{2,i}$ in an indivdual $i$, these could be
say the individual's leg length and nose length. As before we can write
these as 
\begin{eqnarray}
X_{1,i} &= \mu_1+ X_{1,A,i} + X_{1,E,i}  \nonumber \\
X_{2,i} &= \mu_2 +X_{2,A,i} + X_{2,E,i} \nonumber \\
\end{eqnarray}
As before we can talk about the total phenotypic variance ($V_1,V_2$),
environmental variance  ($V_{1,E}$ and $V_{2,E}$), and the additive genetic variance in trait one and two
($V_{1,A}$, $V_{2,A}$). But now we also have to consider the 
total covariance $V_{1,2}=Cov(X_{1},X_{2})$, the environmentally induced covariance between the traits ($V_{E,1,2}=Cov(X_{1,E}
,X_{2,E} )$) and the additive genetic covariance ($V_{A,1,2}
=Cov(X_{1,A} ,X_{2,A} )$) between trait one and two.

We can store these values in a matrices 
\begin{equation}
\bf{V}= \left( \begin{array}{cc} 
V_{1} & V_{1,2} \\
V_{1,2} & V_{2} \\
\end{array} \right) \label{P_matrix}
\end{equation}
and
\begin{equation}
\bf{G}= \left( \begin{array}{cc} 
V_{1,A} & V_{A,1,2} \\
V_{A,1,2} & V_{2,A} \\
\end{array} \right)  \label{G_matrix}
\end{equation}
we can generalize this to an abitrary number of traits.

We can estimate these quantities, in a similar way to before, by
studying the covariance in different traits between relatives: 

\begin{equation}
Cov(X_{1,i},X_{2,j}) = 2 F_{i,j} V_{A,1,2}
\end{equation}

\paragraph{The response of multiple traits to selection, the
  multivariate breeder's equation.}
We can generalize these results for multiple traits, to ask how selection on
multiple phenotypes plays out over short time intervals. We'll write
our change in the mean our multiple phenotypes within a generation as
the vector $\bf{S}$ and our response across multiple generations as
the vector $\bf{R}$. These two quantities are related by 
\begin{equation}
\bf{R} = \bf{G} \bf{V}^{-1} \bf{S} = \bf{G} \bf{\beta}
\end{equation}
 where $\bf{V}$ and $\bf{G}$ are our matrices of the
 variance-covariance of phenotypes and additive genetic values
 (eqn. \eqref{G_matrix} \eqref{P_matrix}) and
 $\bf{\beta}$ is a vector of selection gradients (i.e. the change
 within a generation as a fraction of the total phenotypic variance).
To make this a bit more intuitive, consider two traits we are writing 
\begin{eqnarray}
R_1 & = V_{A,1} \beta_1 + V_{A,1,2} \beta_2 \nonumber \\
R_2 & = V_{A,2} \beta_2 + V_{A,1,2} \beta_1  \nonumber \\
\end{eqnarray}
where the $1$ and $2$ index our two different traits. This is a
statement that our response in any one phenotype is modified by
selection on other traits that covary with that trait.
This offers a good way to think about how genetic trade offs play out
over evolution over short time-scales.

\newpage
\subsection{Non-additive variation.}

Up to now we've assumed that our alleles contribute to our phenotype in an
additive fashion. However, that does not have to be the case as there may be
non-additivity among the alleles present at a locus (dominance) or among
alleles at different loci (epistasis). We can accommodate these complications
into our models. We do this by partitioning our total genetic variance into
independent variance components.

Consider autosomal biallelic locus $\ell$, with frequency $p$ for allele 1, and
genotypes $0$, $1$, and $2$ corresponding to how many copies of allele
1 individuals carry. We'll denote the mean phenotype of an individual
with genotype $0$, $1$, and $2$ are $\overline{X}_{\ell,0}$,
$\overline{X}_{\ell,1}$, $\overline{X}_{\ell,2}$ respectively. Here this mean is
taken over all the environments and genetic backgrounds the alleles
are present on. We'll mean center (MC)
these phenotypic values setting $\overline{X}'_{\ell,0} = \overline{X}_{\ell,0} - \mu$, and
likewise for the other genotypes. 

To illustrate the approach we'll plot two different cases of dominance
relationship in Figure \ref{fig:add_dom}. In the first row of Figure
\ref{fig:add_dom} we show the relationship between genotype and MC phenotype
under an additive model, and in the second row under a model where the allele 1
is dominant, such that the phenotype of the genotype of the heterozyote is the
same as the $11$ homozygote. The area of each circle is proportion to the
fraction of the population in each genotypic class ($p^2$, $2pq$, and $q^2$). 

\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{figures/additive_effect.pdf}
\end{center}
\caption{The average mean-centered (MC) phenotypes of each genotype. 
{\bf Top Row:} Additive relationship between genotype and phenotype. 
{\bf Bottom Row:} Allele 1 is dominant over allele 2, such that the
heterozygote has the same phenotype as the $22$ genotype ($2$). 
The area of each circle is proportion to the fraction of
the population in each genotypic class ($p^2$, $2pq$, and $q^2$). 
One the left column $p=0.1$ and the right column is $p=0.9$.
The additive genetic values of the genotypes are shown as
  red dots. The regression between phenotype and additive genotype is
  shown as a red line. The black vertical arrows show the difference
between the average MC phenotype and additive genetic value for each genotype. } \label{fig:add_dom}
\end{figure}

The first variance component is the variance due to
the additive contribution of each allele ($V_A$). 
We can think about the average
(marginal) MC
phenotype for an allele 1 as the average of the MC phenotype
for heterozgotes and 11 homozygotes weighted by the probability that
an allele $1$ in present in these genotypes. 
These marginal MC
phenotypes for allele 1 and 2 are given by 
\begin{equation} 
a_{\ell, 1} = p\overline{X}'_{\ell,2}  + q\overline{X}'_{\ell,1}, ~~ a_{\ell, 2} = p\overline{X}'_{\ell,1}  + q\overline{X}'_{\ell,0} 
\end{equation}
the marginal value for allele 1, $a_{\ell, 1}$,  follows from the fact that (assuming HW) an allele 1 will be
paired with another allele 1 with probability $p$, resulting in a
genotype $11$ (with phenotypic deviation $\overline{X}'_{\ell,2}$) and will
be paired with an allele 2 with probability $q$ in a heterozygote
(with phenotypic deviation $\overline{X}'_{\ell,1}$). A similar argument
can be made for $a_{\ell, 2}$. \\

The additive MC genetic values (breeding values) of genotype 0, 1, and
2 are then 
\begin{center}
\begin{tabular}{cccc}
genotype: & 0, & 1, & 2.\\
additive genetic value: & $a_{\ell,2}+ a_{\ell,2}$, & $a_{\ell,1}+a_{\ell,2}$, & $a_{\ell,1}+a_{\ell,1}$   \label{add_values}
\end{tabular}
\end{center}
%
Here we are simply adding up the additive contributions of the alleles present
in each genotype. These are the genotypic values of the trait that would result
taking only the additive component of the genotype.  They also have the
interpretation as being the mean phenotype of each genotypes' offspring
averaged over across all possible matings to other individuals in the
population (assuming individuals mate at random). 

The additive genetic values of the genotypes are shown as red dots in Figure
\ref{fig:add_dom}. Note that the additive values of the genotypes line up with
the observed MC phenotypic means in the top row when our alleles interact in a
completely additive manner. Our additive genetic values always fall along a
linear line (the red line in our figure). The additive values are falling best
fitting line of linear regression for our population, when phenotype is
regressed against the additive genotype ($0$, $1$, $2$ copies of allele 1)
across all individuals in our population. Note in the dominant case the
additive genetic values differ from the observed phenotypic means, and are
closer to the observed values for the genotypes that are common in the
population. \\

The difference in the additive effect of the two alleles $a_{\ell, 2}-a_{\ell,
1}$ can be interpreted as a average effect of swapping an allele 1 for an
allele 2, we'll call this difference $\alpha_{\ell}=a_{\ell, 2}-a_{\ell, 1}$.
Our $\alpha_{\ell}$ is also the slope of the regression of genotype against
phenotype (the red line in Figure \ref{fig:add_dom}). Note that the slope of
our regression of genotype on phenotype ($\alpha_{\ell}$) does not depend on
allele frequency for our completely additive locus (top row of
\ref{fig:add_dom}). In contrast, when there is dominance, our the slope between
genotype on phenotype ($\alpha_{\ell}$) is a function of allele frequency
(bottom row of \ref{fig:add_dom}). When a dominant allele (1) is rare there is
a strong slope of genotype on phenotype, bottom left Figure \ref{fig:add_dom}.
This strong slope is because replacing a single copy of the 2 allele with a 1
allele in an individual has a big effect on average phenotype, as it will most
likely move an individual from being a 22 homozygote to being a 12
heterozygote. However, when the dominant allele (1) is common in the
population, replacing a 2 allele by a 1 allele in an individual on average has
little phenotypic effect. This small effect is because as we are mainly turning
heterozygotes into homozygotes (11), who have the same mean phenotype.  \\

The variance in the population phenotype due to these
additive breeding values at locus $\ell$ assuming HW proportions is
\begin{align}
V_{A, \ell} &= p^2 (2a_{\ell,2})^2 + 2pq (a_{\ell,1}+a_{\ell,1})^2 + q^2
(2a_{\ell,0})^2 \nonumber \\
& = 2(p a_{\ell, 1}^2 + q a_{\ell, 2}^2 ) \nonumber \\
& = 2pq \alpha_{\ell}^2
\end{align}
The total additive effect can
be found by summing this additive genetic variance across loci
\begin{equation}
V_A = \sum_{\ell=1}^{L} V_{A, \ell} = \sum_{\ell=1}^{L}
2p_{\ell}q_{\ell} \alpha_{\ell}^2.
\end{equation}


Having assigned the additive genetic variance to be the variance
explained by the additive contribution of the alleles at a locus, we
define the dominance variance as the population variance among
genotypes at a locus due to their deviation from additivity.
We can calculate how much each genotypic mean deviates away from its
additive prediction at locus $\ell$ (the length of the arrows in
Figure \label{fig:add_dom}), for example the heterozygote deviates 
\begin{equation}
d_{\ell,1} =\overline{X}'_{\ell,1}  - (a_{\ell,1}+ a_{\ell,2})
\end{equation}
away from its additive genetic value, with similar expressions for
each of the homozygotes ($d_{\ell,0}$ and $d_{\ell,2}$). We can then write the dominance variance at
our locus as the genotype-frequency weighted sum of our squared
dominance deviations
\begin{equation}
V_{D,\ell} = p^2 d_{\ell,0}^2+ 2pq d_{\ell,1}^2+ q^2 d_{\ell,2}^2.
\end{equation}
Writing our total dominance variance as the sum across loci 
\begin{equation}
V_D = \sum_{\ell=1}^{L}  V_{D,\ell}. 
\end{equation}
Having now partitioned all of the genetic variance into additive and
dominant terms, we can write our total genetic variance as 
\begin{equation}
V_{G} = V_A+V_D.
\end{equation}
We can do this because by construction the covariance between our
additive and dominant deviations for the genotypes is zero. We can
define the narrow sense heritability as before
$h^2=V_A/V_P=V_A/(V_G+V_E)$, which is the proportion of phenotypic
variance due to additive genetic variance. We can also define the 
total proportion of the phenotype variance due to genetic differences
among individuals, as the broad-sense heritability $H^2 =
V_G/(V_G+V_E)$. \\

\begin{table}
\begin{center}
\begin{tabular}{| l | c|}
\hline
Relationship (i,j)$^{*}$ &  $Cov(X_i,X_j)$  \\
\hline
parent--child & $\nicefrac{1}{2} V_A$\\
full siblings &$\nicefrac{1}{2} V_A +\nicefrac{1}{4} V_D$\\
identical (monzygotic) twins & $V_A+V_D$ \\
$1^{st}$ cousins & $\nicefrac{1}{8} V_A$\\
\hline
\end{tabular}
\end{center}
\caption{Phenotypic covariance between some pairs of relatives,
  include the dominance variation. $^{*}$Assuming this is the only relationship
the pair of individuals share (above that expected from randomly
sampling individuals from the population). } % doesn't this implicitly assume an infinite population?
\label{table:domcovar}
\end{table}

When dominance is present in the loci influencing our trait ($V_D>0$), we need to modify our
phenotype covariance among relatives to account for this
non-additivity. Specifically our equation for the covariance among a
general pair of relatives
(eqn. \ref{additive_covar_general_rellys} for additive variation) becomes
\begin{equation}
 Cov(X_1,X_2) = 2 F_{1,2} V_A + r_2 V_D
\end{equation}
where $r_2$ is the probability that the pair of individuals share 2
alleles identical by descent, making the same assumptions (other than additivity) that we made in deriving
eqn. \ref{additive_covar_general_rellys}.  In table
\ref{table:domcovar} we show the phenotypic covariance for some common
pairs of relatives. The regression of offspring phenotype on parental
midpoint still has a slope $V_A/V_P$. 

Full sibs and parent-offspring have the same
covariance if there is no dominance variance (as they have the same
kinship coefficient $F_{1,2}$). However, when dominance
is present ($V_D>0$), full-sibs resemble each other more than
parent-offspring pairs. That's because parents and offspring share
precisely one allele, while full-sibs can share both alleles (i.e. the
full genotype at a locus) identical by descent. We can attempt to
estimate $V_D$ by comparing different sets of relationships. For
example non-identical twins (full sibs born at same time) 
should have $1/2$ the phenotypic covariance of identical twins if
$V_D=0$. Therefore, we can attempt to estimate $V_D$ by looking at
whether identical twins have more than twice the phenotypic covariance
than non-identical twins. \\

The most important aspect of this discussion for thinking about
evolutionary genetics is that the parent-offspring covariance is still
only a function of $V_A$. This is because our parent (e.g. the mother) transmits only a
single allele, at each locus, to its offspring. The other allele the
offspring receives is random (assuming random mating), as it comes
from the other unrelated parent (the father). Therefore, the average
effect on the offspring phenotype of
the allele the offspring receives from the mother, is averaged over
the random allele the child receives from the father. Thus we only
care about the additive effect of the allele, as parents transmit only
alleles not genotypes to their offspring. This means that the short-term response
to selection, as described by the breeder equation, depends only on
$V_A$ and the additive effect of alleles. Therefore, if we can
estimate the narrow-sense heritability we can predict the short-term response.
However, if alleles display dominance our value of $V_A$ will change
as our loci change frequency, e.g. as dominant alleles become common
in the population their contribution to $V_A$ decreases. Therefore,
our value of $V_A$ will not be a constant across generations.

Up to this point we have only considered
dominance and not epistasis. However, we include epistasis in a
similar manner (for example among pairs of loci). This gets a little
tricky to think about so we will only briefly explain it. 
 We can first estimate the additive effect of the
alleles by considering the effect of the alleles averaging over the
other alleles they are paired with, just as before. We can then
calculate the additive genetic variance from this. We can estimate
the dominance variance, by calculating the residual variance among
genotypes at a locus unexplained by the additive effect of the
loci. We can then estimate the epistatic variance by estimating the
residual variance left unexplained among the two locus genotypes from the additive and dominant
deviations calculated from each locus separately. In practice these
high variance components are hard to estimate, and usually small as
much of our variance is assigned to the additive effect.  Again we
would find that we mostly care about $V_A$ for predicting short-term
evolution, but that the contribution of loci to the additive genetic
variance will depend on the epstatic relationships among loci.


\begin{question}
You collect observations of red deer within a generation; recording an
individual’s number of offspring and phenotypes, which are known to
have additive genetic variation, and construct the plots shown in
Figure \ref{fig:red_deer_Q} (standardizing the phenotypes). Answer the following
questions. For each question choose one of the bold options. Briefly justify each of your answers with reference to the breeder's
equation and multi-trait breeder's equation. No calculations are required. \\
{\bf A)}	Looking at just at figure \ref{fig:red_deer_Q} A, in what direction do you expect male antler size to evolve? \\
{\bf Insufficient information, increase, decrease.}\\

{\bf B)}	Looking just at figures \ref{fig:red_deer_Q} B and C, in what direction do you expect male antler size to evolve? \\
{\bf Insufficient information, increase, decrease.}\\

{\bf C)}	(3 Points) Looking at figures \ref{fig:red_deer_Q} A, B, and C in what direction do you expect male antler size to evolve? \\
{\bf Insufficient information, increase, decrease.}\\
\end{question}

\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{figures/Red_deer_selection.pdf}
\end{center}
\caption{ Observations of red deer within a generation; recording an
individual’s number of offspring and phenotypes, which are known to
have additive genetic variation. The figures left to right are A-C.} \label{fig:red_deer_Q}
\end{figure}



\begin{question}
How could you use 1/2 sibs vs full-sibs to estimate $V_D$? Why might
this be difficult in practice? Why are identical vs non-identical
twins better suited for this?
\end{question}

\begin{question}
Can you construct a case where $V_A=0$ and $V_D>0$? You need
just describe it qualitatively, you don't need to work out the
math. (tricker question).
\end{question}



\newpage