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Effects of Non-Random Mating

The models that have been discussed in this course have assumed an infinitesimal animal model, and that animals are mating randomly. What are the effects when these assumptions are not true.

Non-random mating produces the following consequences.

1.

Causes changes to gene frequencies at all loci.

2.

Changes in gene frequencies cause a change in genetic variance. Recall that

$$\sigma^{2}_{G} = 2pq[a+d(q-p)]^{2}+[2pqd]^{2}.$$

3.

In finite populations, non-random mating causes a reduction in the effective population size, which subsequently causes an increase in levels of inbreeding.

4.

Joint equilibrium becomes joint disequilibrium, and therefore, non-zero covariances between additive and dominance genetic effects are created.

5.

If the pedigrees of animals are not complete nor traceable to the base generation, then non-random mating causes genetic evaluations by BLUP to be biased, and causes estimates of genetic variances, by any method, to be biased.

6.

Because the genetic variance decreases due to joint disequilibrium and inbreeding, response to selection is generally lower than expected by selection index over several generations.

1. Effect on Inbreeding

Recall that under random mating and finite population size that the change in inbreeding per generation was defined as

$$\Delta F = \frac{1}{2N_{e}},$$

where N_e is the effective population size, or effective number of breeding individuals. Wright (1940) and Crow (1954) showed that if the average family size, was v, then

N_e = 4N/(v+2),

where N is the number of mating males and females. An idealized population that is able to maintain itself has v=2, thus, N_e=N. Belonsky and Kennedy (1985) simulated a population of 100 females and 5 males per year over ten generations with a single record per animal and discrete generations. Parents were either randomly selected, selected on the basis of phenotype, or selected on the basis of BLUP EBVs. The change in inbreeding under the different selection criteria are given in the table below.

Change in Inbreeding

Selection	Heritability
Criteria	.1	.3	.6
Random	.152	.152	.152
Phenotypic	.169	.206	.215
BLUP EBV	.288	.299	.274

Under random selection of mates, the rate of inbreeding over 10 years is the same for all heritabilities, and is lower than the other forms of mate selection. Inbreeding accumulates due simply to the small population size. Phenotypic selection of mates gives higher rates of inbreeding which increase with increasing heritabilities. Selection of mates using BLUP EBV created the highest rates of inbreeding. Note that there was a decline in inbreeding rate at heritability equal to .6. Why? BLUP EBVs make use of information from relatives. An animal model EBV can be written generally as

$$\hat{a}_{i} = \omega_{1}(\verb*/Data/) + \omega_{2} (\verb*/Parent Ave./) + \omega_{3}(\verb*/Progeny Ave./),$$

where

$$\begin{eqnarray*}\omega_{1} & = & \verb*/ ...$$

under the assumption of no inbreeding and both parents being known. The following table shows values of $\omega$'s for 3 heritabilities.

h2	k	D	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$
			Data	Parents	Progeny
.1	9	28	.036	.643	.321
.3	2.3333	8	.125	.583	.292
.6	0.6667	3	.333	.444	.222

At low heritabilities an animal's EBV is more heavily influenced by the parent average, and many progeny are needed to overcome this influence. Selection on the parent average results in more related animals being selected together, and hence more inbreeding. At the higher heritability of .6, the animal's own record has more influence relative to the parent average, resulting in fewer half-sibs and full-sibs being selected as parents of the next generation, and consequently lower inbreeding levels. Longyang Wu (2000) explored different methods to reduce the influence of the parent average in BLUP EBVs for animals with the aim to reduce inbreeding while not sacrificing on genetic response.

2. Effects on Variances

Assume a normal distribution and truncation selection (i.e. selecting only the higher values), then the mean of the selected group is

$$\mu_{s} = \mu + i \sigma,$$

where $\sigma$ is the standard deviation of the normal distribution, and i is the selection differential, or mean of the selected animals in a normal distribution with mean 0 and variance 1. The truncation point is designated by t and corresponds to p % of the population being selected. Then, the variance of the selected animals is

$$\sigma^{2}_{s} = (1 - k)\sigma^{2},$$

where k=i(i-t).

Traits are correlated, and therefore, selection on one trait can indirectly cause changes to all other traits. Assume two variables, y and x, such that they have a bivariate normal distribution,

$$\left( \begin{array}{c} y \\ x \end{array} \right) \sim N \le... ...\\ \sigma_{xy} & \sigma^{2}_{x} \end{array} \right) \right].$$

The conditional variance of x given y is

$$Var(x \mid y) = \sigma^{2}_{x} - (\sigma_{xy})^{2}/\sigma^{2}_{y}.$$

Using the previous results, then truncation selection on y gives

$$E(y_{s}) = \mu_{y} + i \sigma_{y},$$

and

$$Var(y_{s}) = \sigma^{2}_{y_{s}} = (1-k)\sigma^{2}_{y}.$$

The effects on the correlated variable are

$$E(x \mid y_{s}) = \mu_{x} + b_{x,y} ( y_{s} - E(y_{s})),$$

where

$$b_{x,y} = \frac{\sigma_{yx}}{\sigma^{2}_{y}},$$

is the regression of x on y. Now

$$Var(x_{s}) = Var(x \mid y_{s}) + Var(E(x \mid y_{s})),$$

where

$$\begin{eqnarray*}Var(x \mid y_{s}) & = & \sigma^{2}_{x} - \frac{(\sigma_{yx})^{2... ...c{(\sigma_{yx})^{2}}{(\sigma^{2}_{y})^{2}} (1-k)\sigma^{2}_{y}, \end{eqnarray*}$$

so that

$$\begin{eqnarray*}Var(x_{s}) & = & (1-r^{2}_{yx})\sigma^{2}_{x}+ \frac{(\sigma_{... ...x}] \sigma^{2}_{x}, \\ & = & (1 - kr^{2}_{yx})\sigma^{2}_{x}. \end{eqnarray*}$$

Let

$$Var \left( \begin{array}{c} y \\ x \end{array} \right) = \left( \begin{array}{rr} 100 & -10 \\ -10 & 4 \end{array} \right),$$

so that

$$r_{yx} = \frac{-10}{(10)(2)} = -.5,$$

and r²_yx = .25. If truncation selection to keep the best 10 % is applied, then i=1.755, t=1.2821, and k=.83. Then the variances after selection would be

$$\begin{eqnarray*}\sigma^{2}_{y_{s}} & = & (1-.83)\sigma^{2}_{y} = 17, \\ \sigma... ...\ \sigma^{2}_{x_{s}} & = & (1-(.83)(.25))\sigma^{2}_{x} = 3.17. \end{eqnarray*}$$

3. Effect on the Animal Model

Using the previous results, selection is on the phenotypic values,

$$y_{i} = \mu + a_{i} + e_{i},$$

and suppose that x_i = a_i, then

$$\begin{eqnarray*}\sigma_{yx} & = & \sigma_{ya} = \sigma^{2}_{a}, \\ \sigma^{2}_... ...})^{2}}{\sigma^{2}_{y}} k \\ & = & (1-k \ h^{2})\sigma^{2}_{a} \end{eqnarray*}$$

Similarly, if x_i = e_i, then

$$\begin{eqnarray*}\sigma_{yx} & = & \sigma_{ye} = \sigma^{2}_{e}, \\ \sigma^{2}_... ...} {\sigma^{2}_{y}} \\ & = & (1 - k \ (1-h^{2})) \sigma^{2}_{e} \end{eqnarray*}$$

Let $\sigma^{2}_{y}=700$, $\sigma^{2}_{a}=210$, and $\sigma^{2}_{e}=490$, then h²=.3. Assume 50% selection intensity, which gives i=.8, t=0, and k=.64, then

$$\begin{eqnarray*}\sigma^{2}_{y_{s}} & = & (1-.64)(700) = 252, \\ \sigma^{2}_{a_... ....68, \\ \sigma^{2}_{e_{s}} & = & (1-(.64)(1-.3))(490) = 270.48. \end{eqnarray*}$$

Note that because of selection there is a non-zero covariance between a_s and e_s, so that

$$\sigma^{2}_{y_{s}} = \sigma^{2}_{a_{s}} + \sigma^{2}_{e_{s}} + 2 \sigma_{a_{s}e_{s}},$$

and given the above variances, the implied covariance is

$$\sigma_{a_{s}e_{s}} = .5(252 - 169.68 - 270.48) \ = \ -94.08.$$

3.1 The Next Generation

Animals that were included in y_s are allowed to randomly mate with each other to give a progeny generation. What happens to their variances and covariances? The variance of residual effects should be equal to the residual variance of the original population because selection should not affect future residual effects. Thus,

$$\sigma^{2}_{e_{1}} = \sigma^{2}_{e} = 490.$$

However, the additive genetic variance will depend upon the selection that has occurred, because there is less additive genetic variance in the selected parents. An assumption is that the Mendelian sampling variance is not affected by selection (but could be reduced due to inbreeding). Then

$$\begin{eqnarray*}\sigma^{2}_{a_{1}} & = & .25 \sigma^{2}_{a_{s}}({\rm males}) + ... ...a^{2}_{a} \\ & = & (1 - .5 (.64) (.3))(210) \\ & = & 189.84. \end{eqnarray*}$$

Then it follows that

$$\sigma^{2}_{y_{1}} = \sigma^{2}_{a_{1}} + \sigma^{2}_{e_{1}} = 189.84 + 490 = 679.84,$$

or

$$\sigma^{2}_{y_{1}} = (1 - .5 h^{4}k)\sigma^{2}_{y}.$$

Another generation of random mating would produce the following results.

$$\begin{eqnarray*}\sigma^{2}_{e_{2}} & = & 490, \\ \sigma^{2}_{a_{2}} & = & .5 \... ...64))210 \\ & = & 199.92, \\ \sigma^{2}_{y_{2}} & = & 689.92. \end{eqnarray*}$$

Continued random matings of progeny in subsequent generations would re-generate the lost genetic variance due to the original selection. With small populations, the Mendelian sampling variance would decrease due to inbreeding, and therefore, the original amount of additive genetic variance would never be achieved.

3.2 Another Cycle of Selection

Suppose that 50% truncation selection had been applied to y₁as in the original population, prior to matings. The heritability in generation 1 is

$$h^{2}_{1} = \sigma^{2}_{a_{1}}/ \sigma^{2}_{y_{1}},$$

or h²₁ = .27924 = 189.84/679.84. Thus, heritability has decreased as a result of selection. The additive genetic variance of the selected animals from generation 1 is

$$\begin{eqnarray*}\sigma^{2}_{a_{1s}} & = & (1 - h^{2}_{1}k) \sigma^{2}_{a_{1}} \\ & = & (1 - (.27924)(.64))(189.84) \\ & = & 155.91 \end{eqnarray*}$$

Now the variances of the progeny (generation 2) from random mating of the selected generation 1 animals is

$$\begin{eqnarray*}\sigma^{2}_{a_{2}} & = & .5 \sigma^{2}_{a_{1s}} + .5 \sigma^{2}... ...h^{2}_{1}k \sigma^{2}_{a_{1}} \\ & = & \sigma^{2}_{a} - d_{2}, \end{eqnarray*}$$

where d₁ is the disequilibrium generated by the selection on the original population, and d₂ is the new disequilibrium generated by selection on generation 1. Note that

$$d_{2} = .5 d_{1} + .5 h^{2}_{1}k \sigma^{2}_{a_{1}},$$

which is one half the 'old' disequilibrium plus a 'new' disequilibrium. This recursion process can be used to deduce the limiting value of the disequilibrium. At generation t,

$$\begin{eqnarray*}\sigma^{2}_{a_{t}} & = & \sigma^{2}_{a} - d_{t} \\ d_{t+1} & = & .5 d_{t} + .5 h^{2}_{t}k \sigma^{2}_{a_{t}}, \end{eqnarray*}$$

assuming the same selection intensity in each generation. Equate d_t+1 to d_t, then

$$\begin{eqnarray*}d & = & .5 d + .5 h^{2}_{t}k \sigma^{2}_{a_{t}} \\ d & = & h^{2}_{*} k \sigma^{2}_{a_{*}}. \end{eqnarray*}$$

An equilibrium is attained when the new disequilibrium equals one half the old disequilibrium. All of the above development assumed an infinite population size so that the effects of inbreeding could be ignored.

3.3 Kennedy studies

Kennedy (1985) calculated response to selection for repeated cycles of selection for 4 generations with a selection intensity of 20%, which gives k=.7818 and i=1.4. The starting heritability was .5, with a phenotypic variance of 100. The response to selection was

$$R = i \ h^{2}_{t} \sigma_{y_{t}}.$$

Typically, response to selection is assumed to be the same in each generation. Using the base population values the response after one generation would be

R = ( 1.4 (.5) (10)) = 7.00.

After 4 generations of selection, the total response would be 28.00. However, this is an overestimate because it does not account for the decrease in variance due to joint disequilibrium. Below is the table generated by Kennedy showing the generation by generation response.

Generation	$\sigma^{2}_{y_{t}}$	h2t	dt	R
0	100	.5	0	-
1	90.2	.446	-9.8	7.00
2	88.1	.432	-11.9	5.93
3	87.6	.429	-12.4	5.68
4	87.5	.428	-12.5	5.63
$\infty$	87.5	.428	-12.5	5.61

The total response over four generations is only

$$R = 7.00 \ + \ 5.93 \ + \ 5.68 \ + \ 5.63 \ = \ 24.24,$$

which is 84.5% of the expected response of 28. Note that the balance in old and new equilibrium was achieved very quickly in only 4-5 generations.

Many studies estimated the realized heritability from these kinds of selection experiments, by dividing the accumulated response by the accumulated selection differential.

Generation	$R= ih^{2}\sigma_{y}$	$S= i\sigma_{y}$	Realized	Actual
			h2	h2
0	7.00	14.00	.500	.500
1	12.93	27.30	.474	.446
2	18.61	40.44	.460	.432
3	24.24	53.53	.453	.429

Obviously, realized heritability is not a good estimate of the heritability in any generation, except the initial generation. Instead of using accumulated response and selection differentials, the change from one generation to the next will give an estimate of the actual heritability.

3.4 Effect on Genetic Evaluation

The effects of non-random mating on genetic evaluation are minimal if

• Complete (no missing parent information) pedigrees are known back to a common base population which was mating randomly,
• Data on all candidates for selection are available, and
• Genetic parameters from the base population are known.

If the above conditions hold, then application of BLUP does not lead to bias in EBVs, but selection does increase the variance of prediction error over populations that are randomly mating. However, in animal breeding, the practical situation is that complete pedigrees seldom exist. Thus, bias can creep into estimates of fixed effects and EBVs.

Recall that HMME for a simple animal model are

$$\left( \begin{array}{ccc} {\bf X}'{\bf R}^{-1}{\bf X} & {\bf... ...\ {\bf Z}'{\bf R}^{-1}{\bf y} \\ {\bf0} \end{array} \right),$$

where $k_{a} = \sigma^{-2}_{a}$. A generalized inverse of the coefficient matrix can be represented as

$$\left( \begin{array}{ccc} {\bf C}_{xx} & {\bf C}_{xn} & {\bf... ...\bf C}_{ox} & {\bf C}_{on} & {\bf C}_{oo} \end{array} \right).$$

Then remember that

$$Var \left( \begin{array}{c} \hat{\bf a}_{n} - {\bf a}_{n} \\ ... ... C}_{no} \\ {\bf C}_{on} & {\bf C}_{oo} \end{array} \right),$$

and that

$$\begin{eqnarray*}Cov(\hat{\bf b},\hat{\bf a}_{n}) & = & {\bf0}, \\ Cov(\hat{\bf b},{\bf a}_{n}) & = & -{\bf C}_{xn}. \end{eqnarray*}$$

These results indicate that HMME forces the covariance between estimates of the fixed effects and estimates of additive genetic effects to be null. However, there is a non-zero covariance between estimates of the fixed effects and the true additive genetic values of animals. Hence, any problem with the true additive genetic values, and there will be problems with estimates of fixed effects.

Consider the equation for $\hat{\bf b}$,

$$\hat{\bf b} = ({\bf X}'{\bf R}^{-1}{\bf X})^{-} ({\bf X}'{\bf R}^{-1}{\bf y} - {\bf X}'{\bf R}^{-1}{\bf Z} \hat{\bf a}_{n}),$$

and the expectation of this vector is

$$E(\hat{\bf b}) = ({\bf X}'{\bf R}^{-1}{\bf X})^{-} ({\bf X}'... ...1}{\bf Xb} - {\bf X}'{\bf R}^{-1}{\bf Z} E(\hat{\bf a}_{n})).$$

The fixed effects solution vector contains a function of the expectation of the additive genetic solution vector. Normally, because the BLUP methodology requires

$$E(\hat{\bf a}_{n}) = E({\bf a}_{n}) = {\bf0},$$

then the fixed effects solution vector is also unbiased. Due to selection, however,

$$E({\bf a}_{n}) \neq {\bf0},$$

and therefore, the expectation of the fixed effects solution vector contains a function of $E({\bf a}_{n})$ and is consequently biased. If $\hat{\bf b}$ is biased, then this will cause a bias in $\hat{\bf a}$.

3.5 Alternative Method

Re-state the model (in general terms) as

$${\bf y} = {\bf Xb} + {\bf Zu} + {\bf e},$$

where

$$E \left( \begin{array}{c} {\bf u} \\ {\bf e} \end{array} \rig... ...\left( \begin{array}{c} {\bf u} \\ {\bf0} \end{array} \right),$$

and therefore,

$$E({\bf y}) = {\bf Xb}+{\bf Zu}.$$

To simplify, assume that ${\bf G}=Var({\bf u})$ and ${\bf R}=Var({\bf e})$ and that neither is drastically affected by non-random mating.

The prediction problem is the same as before. Predict a function of ${\bf K'}{\bf b}+{\bf M}'{\bf u}$ by a linear function of the observation vector, ${\bf L}'{\bf y}$, such that

$$\begin{eqnarray*}E({\bf K'}{\bf b}+{\bf M}'{\bf u}) = E({\bf L}'{\bf y}), \end{eqnarray*}$$

and such that $Var({\bf K'}{\bf b}+{\bf M}'{\bf u}-{\bf L}'{\bf y})$is minimized. Form the variance of prediction errors and add a LaGrange multiplier to ensure the unbiasedness condition, then differentiate with respect to the unknown ${\bf L}$ and the matrix of LaGrange multipliers and equate to zero. The solution gives the following equations.

$$\left( \begin{array}{cc} {\bf X}'{\bf V}^{-1}{\bf X} & {\bf X... ...1}{\bf y} \\ {\bf Z}'{\bf V}^{-1}{\bf y} \end{array} \right).$$

Because ${\bf V} = {\bf ZGZ}'+{\bf R}$, and

$${\bf V}^{-1} = {\bf R}^{-1}-{\bf R}^{-1}{\bf ZTZ}'{\bf R}^{-1},$$

for ${\bf T}=({\bf Z}'{\bf R}^{-1}{\bf Z}+{\bf G}^{-1})^{-1}$, then it can be shown that the following equations give the exact same solutions as the previous equations.

$$\left( \begin{array}{cc} {\bf X}'{\bf R}^{-1}{\bf X} & {\bf X... ...1}{\bf y} \\ {\bf Z}'{\bf R}^{-1}{\bf y} \end{array} \right).$$

If a generalized inverse to the above coefficient matrix is represented as

$$\left( \begin{array}{cc} {\bf C}_{xx} & {\bf C}_{xz} \\ {\bf C}_{zx} & {\bf C}_{zz} \end{array} \right),$$

then some properties of these equations are

$$\begin{eqnarray*}Cov(\hat{\bf b},{\bf u}) & = & {\bf0}, \\ E(\hat{\bf b}) & = &... ...1}{\bf Xb}, \\ Cov(\hat{\bf b},\hat{\bf u}) & = & {\bf C}_{xz}. \end{eqnarray*}$$

Firstly, these results suggest that if non-random mating has occurred and has changed the expectation of the random vector, then an appropriate set of equations is the generalized least squares equations. However, we have seen earlier that such equations give a lower correlation with true values and large mean squared errors (when matings are at random). Secondly, the estimates of the fixed effects have null covariances with the true random effects, and the covariances between estimates of the fixed effects and estimates of the random effects are non-zero, which is opposite to the results from BLUP. With the least squares solutions, application of the regressed least squares procedure could be subsequently used to give EBVs.

There is another problem with these equations. If ${\bf u}={\bf a}$as in an animal model, then ${\bf Z}={\bf I}$, and the generalized least squares equations do not have a solution unless $\hat{\bf a}= {\bf0}$. This is not very useful for genetic evaluation purposes.

4. Genetic Response

The simple way to estimate genetic response to selection is to average the EBVs from an animal model by year of birth for males and females, separately.

Kennedy did another simulation to show that when full pedigree information is known and all data are used with the correct heritability in generation 0, that MME lead to unbiased estimates of genetic response. He generated 160 replicates of selecting 5 males out of 20 each generation for a trait with an initial heritability of .5 and genetic variance of 10. He looked at both random selection of males and truncation selection of males (based on phenotypes).

Gen.	Random			Truncation
	$\sigma^{2}_{a}$	true R	MME R	$\sigma^{2}_{a}$	true R	MME R
1	9.9	20.10	19.91	9.82	20.02	19.98
2	9.5	20.12	20.16	8.56	21.43	21.37
3	9.2	19.89	19.95	8.39	22.64	22.63

The decrease in genetic variance under random selection was due solely to the accumulation of inbreeding, but the decrease under truncation selection was due to inbreeding and joint disequilibrium.

Kennedy also studied what would happen if an incorrect heritability value was used in MME, using the same simulation strategy as above.

Gen.	Random	Truncation
	h2=.7	h2=.3	h2=.7
1	.09	.27	-.13
2	-.02	-.09	.12
3	-.08	-.19	.24

Under random selection, there was no bias in estimated genetic response, but under truncation selection an underestimation of genetic response occurred when a smaller than true heritability was used (i.e. 0.3), and an overestimation occurred with a larger than true heritability (i.e. 0.7). Wu(2000) also made a similar study as Kennedy, but did 200 replicates over 20 generations of selection with truncation selection on both males and females using MME EBVs. Wu(2000) deliberately used higher heritability values in MME in order to reduce inbreeding in the population, and over a longer term of 20 generations the higher heritability value gave greater true genetic response because genetic variability was not reduced. The initial phenotypic variance was 100.

True h2	h2 used	Response	Inbreed.	$\sigma^{2}_{a}$
	in MME
.1	.1	6.26	.74	2.45
	.5	6.76	.57	4.01
.25	.25	9.20	.71	7.03
	.50	9.74	.56	10.18
.5	.50	11.58	.63	15.56
	.67	11.79	.58	17.38

Gasarabwe(2001) obtained similar results to Wu(2000) in a simulation of a closed swine nucleus herd of different sizes from 200 to 500 sows per generation and different numbers of sows per boar mating structure. Using the larger heritabilities gave reduced inbreeding and greater genetic responses.

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