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Covariances Between Relatives

1. Parent-Offspring

Progeny will inherit one-half the additive breeding value of the parent. Below is a table that summarizes the quantities that will be needed to calculate the covariance between parent and offspring.

Parent Freq. $G-\mu_{G}$ Parent Offspring
Genotype     BV BV
A1A1 p2 $2q(\alpha-qd)$ $2q\alpha$ $q\alpha$
A1A2 2pq $(q-p)\alpha +2pqd$ $(q-p)\alpha$ $0.5(q-p)\alpha$
A2A2 q2 $-2p(\alpha +pd)$ $-2p\alpha$ $-p\alpha$
The covariance between parent $G-\mu_{G}$ and offspring BV is

\begin{eqnarray*}\sigma_{OP} & = & p^{2}(2q(\alpha-qd))q\alpha \\
& & +2pq[(q-...
... \\
& = & pq \alpha^{2} \\
& = & \frac{1}{2} \sigma^{2}_{A}.

2. Half-Sibs

From the table above, the covariance between half-sibs can be computed. The offspring column represents half-sibs, so that the variance of offspring means gives the covariance between half-sibs.

\begin{eqnarray*}\sigma_{HS} & = & p^{2}q^{2}\alpha^{2} +2pq(\frac{1}{2}(q-p)\al...
... \frac{1}{2} pq \alpha^{2} \\
& = & \frac{1}{4} \sigma^{2}_{A}

3. Midparent - Offspring

The midparent value involves both parents, and the value of the offspring depends on the genotypes of the parents. So the offspring of each possible pair of genotype matings has to be considered.

Possible Parental Frequency Midparent Offspring
Matings   Genotypic Value Mean Value
A1A1 A1A1 p4 a a
A1A1 A1A2 4p3q 0.5(a+d) 0.5(a+d)
A1A1 A2A2 2p2q2 0 d
A1A2 A1A2 4p2q2 d 0.5d
A1A2 A2A2 4pq3 0.5(d-a) 0.5(d-a)
A2A2 A2A2 q4 -a -a
The covariance is obtained by multiplying together the last three columns of the above table and summing, and then subtracting off the population mean squared.

\begin{eqnarray*}\sigma_{O \bar{P}} & = & p^{4}a^{2} + p^{3}q(a+d)^{2} +2p^{2}q^...
...} \\
& = & pq \alpha^{2} \\
& = & \frac{1}{2} \sigma^{2}_{A}

4. Full-Sibs

The table for Midparent-Offspring covariance can be used to obtain the covariance between full-sibs. The offspring within a mating pair are full-sibs, and the variance of offspring means gives the covariance between full-sibs.

\begin{eqnarray*}\sigma_{FS} & = & p^{4}a^{2} + 4p^{3}q(0.5(a+d))^{2} +2p^{2}q^{...
& = & \frac{1}{2} \sigma^{2}_{A} + \frac{1}{4} \sigma^{2}_{D}

5. Phenotypic Values

The phenotype, what is visible, is composed of a mean plus the genotypic value plus a random residual term, i.e.,

\begin{displaymath}P = \mu + G + E. \end{displaymath}

Let a= 1, d=.5, p=.4, $\mu=3$ and $\sigma^{2}_{E} = 1.0$, then a phenotype for an A1A1 individual could be simulated as

\begin{eqnarray*}\mu & = & 3 \\
G & = & 1 \\
E & = & \mbox{ RND }*1.0 \\
P & = & 3 + 1 - .436 \\
& = & 3.564,

where RND indicates a random normal deviate number generator. Assumptions commonly made are that $\mu=0$ and that the average of all residual terms is zero. Then the average phenotypic value of genotype A1A1 is equal to the genotypic value of that genotype, assuming a large population. However, the variance of the phenotypic values will be greater than the variance of genotypic values because the residual variance is greater than zero.

\begin{eqnarray*}\sigma^{2}_{P} & = & \sigma^{2}_{G} + \sigma^{2}_{E} \\
& = &...
...\sigma^{2}_{E} \\
& = & 0.528 + 0.0576 + 1.0 \\
& = & 1.5856

The covariance between genotypic values and residual effects is assumed to be zero.

The phenotypic mean of the population will be

\begin{displaymath}\mu_{P} = \mu + \mu_{G}, \end{displaymath}

where $\mu_{G} = 0.04$ using the above parameters, so that $\mu_{P} = 3.04$. Usually genotypes of animals are unknown, and selection must be based on phenotypes. The rate of genetic change will depend on the size of $\sigma^{2}_{E}$ and values of a and d, and the initial value of p.

Suppose you have two individuals of genotype A1A2 with the male having P = 3 + 0.5 + .621 = 4.121, and the female having P = 3 + 0.5 - .894 = 2.606. To generate a progeny record from mating these two individuals,

Select randomly an allele from the male parent, either A1 or A2, say it is A2.
Select randomly an allele from the female parent in the same manner, say it is also A2.
Generate a new residual effect for the progeny record, say it is -.307, then

P = 3 - 1 - .307 = 1.693.

What would be the expected average phenotype of a large number of progeny from this mating? This mating can produce all three genotypes in the ratio 1:2:1. Those with genotype A1A1 would have a mean of 4, those with A1A2would have a mean of 3.5, and those with A2A2 would have a mean of 2. Weighting these by their expected frequencies, then

\begin{displaymath}\bar{P} = [(4)+(3.5)\times 2+(2)]/4 = 3.25. \end{displaymath}

6. Small Populations and Drift

In large populations gene frequencies remain constant from generation to generation under random mating, no selection, no mutations, and no migration. In small populations there are random changes in gene frequencies due to the sampling of gametes during matings. This change in gene frequency is known as random drift. Hardy-Weinberg equilibrium is lost in a small population.

Within the small population, due to random drift, there is

Assume a large population which is split into subpopulations (lines) of N individuals each. Within each line, assume individuals are randomly mated (including selfing), in discrete generations, and that population size remains constant. The large population is assumed to be in Hardy-Weinberg equilibrium. The allele frequencies in each line are initially assumed to be equal to the population frequencies, p0 and q0.

If there were a large number of lines, then the average frequency of the A2 allele over all lines at generation t is

\begin{displaymath}\bar{q}_{t} = q_{0}. \end{displaymath}

The variation in qt across lines is equal to the drift variance. Assuming two alleles at the locus, then the allele frequencies follow a binomial distribution, and after t generations the drift variance can be shown to be equal to

\begin{displaymath}\sigma^{2}_{q_{t}} = p_{0}q_{0}[1 - (1 - \frac{1}{2N})^{t}]. \end{displaymath}

Below are some values of drift variance for N=100.

q0 t=1 t=5 t=10 t=20 $t=\infty$
.1 .00045 .00223 .00440 .00858 .09
.2 .00080 .00396 .00782 .01526 .16
.3 .00105 .00520 .01027 .02003 .21
.4 .00120 .00594 .01173 .02283 .24
.5 .00125 .00619 .01222 .02385 .25
If q0= .4, then the drift variance was .00120, which means that the gene frequency within lines could vary from .37 to .43 for two thirds of the lines, and from .34 to .46 for 95 % of the lines. After 20 generations, the range would be .25 to .55 for 95 % of the lines. Thus, the changes can be very dramatic within a line.

By definition,

\begin{displaymath}\sigma^{2}_{q_{t}} = E(q^{2}_{t}) - [E(q_{t})]^{2}, \end{displaymath}

and let

\begin{displaymath}\bar{q}_{t}^{2} = E(q^{2}_{t}), \end{displaymath}


\begin{displaymath}E(q^{2}_{t}) = \sigma^{2}_{q_{t}} + [E(q_{t})]^{2}, \end{displaymath}


\begin{displaymath}\bar{q}_{t}^{2} = \sigma^{2}_{q_{t}} + (q_{0})^{2}. \end{displaymath}

Note that $\bar{q}_{t}^{2}$ is the average frequency of A2A2 genotypes over all lines at generation t. On the overall large population basis (all small populations collectively), then

Genotype Frequency  
A1A1 $\bar{p}_{t}^{2}$ $ = p_{0}^{2} + \sigma^{2}_{q_{t}}$
A1A2 $2\bar{p}_{t} \bar{q}_{t}$ $= 2p_{0}q_{0} -
2 \sigma^{2}_{q_{t}}$
A2A2 $\bar{q}_{t}^{2}$ $=q_{0}^{2} + \sigma^{2}_{q_{t}}$
Note that as the number of generations increases, then $\sigma^{2}_{q_{t}}$ also increases towards the limiting value of p0q0. Therefore, eventually there will be no heterozygous genotypes, and all loci within a line will be fixed to either A1A1 or A2A2 with frequency p0 and q0, respectively.

7. Inbreeding

Inbreeding and drift are related phenomena in small populations. Inbreeding results from the mating of related individuals. In a small population the degree of relationship among individuals increases with generation number and depends on the size of the population. Let F represent the coefficient of inbreeding, which is defined as the probability that two genes at a locus are identical by descent.

Consider a population of size N where every parent contributes equally to the next generation. Within the N parents there are 2N possible gametes. If two gametes are chosen at random from the population of 2N gametes, then the probability that the two gametes are identical and from the same parent is $\frac{1}{2N}$. The probability that the two gametes are not from the same parent is therefore, $1 - \frac{1}{2N}$. Even though the two gametes are not from the same parent, it is possible that alleles are equal, for example, both could be A1. The probability that two alleles at a locus are identical by descent is 1 in the first case which occurs with probability $\frac{1}{2N}$. The probability that two alleles at a locus are identical by descent in the second case is Ft-1 in the parent generation, which occurs with probability $1 - \frac{1}{2N}$, so that the coefficient of inbreeding in the progeny generation is

\begin{displaymath}F_{t} = \frac{1}{2N} + (1 - \frac{1}{2N})F_{t-1}. \end{displaymath}

Assume that F0=0, then

\begin{displaymath}F_{1} = \frac{1}{2N}. \end{displaymath}

In the next generation,

\begin{eqnarray*}F_{2} & = & \frac{1}{2N} + (1 - \frac{1}{2N})\frac{1}{2N} \\
& = & 1 - ( 1 - \frac{1}{2N} )^{2}

as a result of completing the square. In general,

\begin{displaymath}F_{t} = 1 - (1 - \frac{1}{2N} )^{t}. \end{displaymath}

In the previous section, drift variance was equal to

\begin{displaymath}\sigma^{2}_{q_{t}} = p_{0}q_{0} [1 - (1 - \frac{1}{2N})^{t}], \end{displaymath}

and therefore,

\begin{displaymath}\sigma^{2}_{q_{t}} = p_{0}q_{0} F_{t}. \end{displaymath}

The drift variance is a function of the coefficient of inbreeding.

Genotype Frequency  
A1A1 $ p_{0}^{2} + \sigma^{2}_{q_{t}}$ p02 + p0q0Ft
A1A2 $ 2p_{0}q_{0} -2 \sigma^{2}_{q_{t}}$ 2p0q0 (1 - Ft)
A2A2 $q_{0}^{2} + \sigma^{2}_{q_{t}}$ q02 + p0q0 Ft
The frequency of heterozygotes goes to zero as the coefficient of inbreeding increases towards 1.

8. Genetic Variance

Assume an additive model for a single locus, so that d=0, then the genetic variance in the base population is

\begin{displaymath}\sigma^{2}_{G_{0}} = 2 p_{0}q_{0} a^{2}. \end{displaymath}

In line i, then

\begin{displaymath}\sigma^{2}_{G_{i}} = 2 p_{i}q_{i} a^{2}. \end{displaymath}

The average within line genetic variance is

\begin{eqnarray*}\sigma^{2}_{G_{w}} & = & 2 \bar{p} \bar{q} a^{2} \\
\bar{p} \bar{q} & = & ( \sum_{i=1}^{l} p_{i}q_{i})/l

Note that $2\bar{p}\bar{q}$ is the overall frequency of heterozygotes across lines, then

\begin{eqnarray*}\sigma^{2}_{G_{w}} & = & 2 p_{0}q_{0} a^{2} (1 - F_{t}) \\
& = & \sigma^{2}_{G_{0}} ( 1 - F_{t}).

The genetic variance between lines is calculated by computing the variance of means of each line. The mean of line i is

Mi = a (pi - qi) = a(1 - 2qi),


\begin{eqnarray*}V(M_{i}) & = & \sigma^{2}_{G_{b}} \\
& = & V[ a (1-2q_{i})] \...
... 4 a^{2} p_{0}q_{0} F_{t} \\
& = & 2 \sigma^{2}_{G_{0}} F_{t}

The total genetic variance is the sum of the within line genetic variance and the between lines genetic variance.

Within Lines $\sigma^{2}_{G_{0}}(1 - F_{t})$
Between Lines $\sigma^{2}_{G_{0}} 2F_{t}$
Total $\sigma^{2}_{G_{0}} ( 1 + F_{t})$
Therefore, as the coefficient of inbreeding, Ft, goes to 1, the within lines genetic variance goes to 0, and the between line genetic variance increases to $2 \sigma^{2}_{G_{0}}$. Within a line the genes become fixed as t approaches $\infty$, and so there would be no genetic variation between individuals.

With a dominance model, $d \neq 0$, then Weir and Cockerham (1977) have shown that the total genetic variance, between and within, is

\begin{displaymath}\sigma^{2}_{G} = (1+F_{t})\sigma^{2}_{A_{0}} + (1 - F_{t})
+ F_{t} \sigma^{2}_{H} - F^{2}_{t} D^{2}_{H}, \end{displaymath}

Cov(A0,D0) is the covariance between additive gene effects and dominance deviations;
$\sigma^{2}_{H}$ is the variance caused by dominance deviations of homozygotes; and
DH is the sum of dominance deviations of homozygotes.
This formula was extended to multiple loci by Cockerham and Weir (1984), Smith and Maki Tanila (1990), and de Boer and Hoeschele (1995).

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This LaTeX document is available as postscript or asAdobe PDF.

Larry Schaeffer