The improvement through breeding

Response to selection

Change, or response ( $R$ ), is given by the basic equation:

$$\begin{align*} R = h^2S \end{align*}$$

• Here $S$ is the selection differential

• Is equal to the mean value of the selected parents ( $\mu_S$ ) minus the population mean ( $\mu$ ).

Response to selection

Change, or response ( $R$ ), is given by the basic equation:

$$\begin{align*} R = h^2S \end{align*}$$

• Here $S$ is the selection differential

• Is equal to the mean value of the selected parents ( $\mu_S$ ) minus the population mean ( $\mu$ ).

The regression of offspring on mid-parent is equal to the heritability

• No non-genetic cause of resemblance
• No natural selection, i.e. same fertility and viability

Therefore,

$$\begin{align*} R = b_{O\bar{P}}S \\ R = h^2S \end{align*}$$

Response to selection

Change, or response ( $R$ ), is given by the basic equation:

$$\begin{align*} R = h^2S \end{align*}$$

Here S is the selection differential, which is equal to the mean value of the selected parents ( $\mu_S$ ) minus the population mean ( $\mu$ ).

Improve the body weight

Response to selection

Change, or response ( $R$ ), is given by the basic equation:

$$\begin{align*} R = h^2S \end{align*}$$

Here S is the selection differential, which is equal to the mean value of the selected parents ( $\mu_S$ ) minus the population mean ( $\mu$ ).

Improve the body weight

Standardized selection differential

With the assumption that the phenotypic distributions are normal.

• Same $S$ means different selection pressure if the variances of the normal distributions are different

Selection differential in standard deviation units:

$$\begin{align*} \frac{S}{\sigma_P} \end{align*}$$

Intensity of selection ( $i$ )

$$\begin{align*} i = \frac{S}{\sigma_P} \end{align*}$$

• $i$ is the "standarized selection differential"

• it better reflects the selection effort

Intensity of selection ( $i$ )

curve(dnorm(x,0,1), xlim=c(-3,3), xaxt="n", xlab="Trait Value", main="", ylab="Density", lwd=3)
fromd <- qnorm(.95); tod <- 3
S.x  <- c(fromd, seq(fromd, tod, 0.01), tod)
S.y  <- c(0, dnorm(seq(fromd, tod, 0.01)), 0)
polygon(S.x,S.y, col="grey")
abline(v=mean(S.x**S.y)*2, col="blue", lwd=3); abline(v=0, col="red", lwd=3)

Intensity of selection ( $i$ )

curve(dnorm(x,0,1), xlim=c(-3,3), xaxt="n", xlab="Trait Value", main="", ylab="Density", lwd=3)
fromd <- qnorm(.95); tod <- 3
S.x  <- c(fromd, seq(fromd, tod, 0.01), tod)
S.y  <- c(0, dnorm(seq(fromd, tod, 0.01)), 0)
polygon(S.x,S.y, col="grey")
abline(v=mean(S.x**S.y)*2, col="blue", lwd=3); 
abline(v=0, col="red", lwd=3)

Intensity of selection ( $i$ )

A table of these values is found in Appendix A of the F&M book.

ifun <- function(p=0.5){
  x=qnorm(p=(1-p)) # get the truncation point
  z=dnorm(x) # get z
  return(z/p) # get i
}
p <- seq(0.005, 1, by=0.005)
i <- ifun(p)
plot(p, i)

head(data.frame(p, i), 20)

##        p        i
## 1  0.005 2.891949
## 2  0.010 2.665214
## 3  0.015 2.524695
## 4  0.020 2.420907
## 5  0.025 2.337803
## 6  0.030 2.268065
## 7  0.035 2.207716
## 8  0.040 2.154344
## 9  0.045 2.106373
## 10 0.050 2.062713
## 11 0.055 2.022578
## 12 0.060 1.985383
## 13 0.065 1.950678
## 14 0.070 1.918113
## 15 0.075 1.887406
## 16 0.080 1.858328
## 17 0.085 1.830692
## 18 0.090 1.804340
## 19 0.095 1.779142
## 20 0.100 1.754983

Breeder's equation

Given: $$\begin{align*} &R = h^2S\\ &S = i\sigma_P& \end{align*}$$

So now, we have, as a prediction:

$$\begin{align*} &R = h^2i\sigma_P\\ \end{align*}$$

where $i = \frac{z}{p}$.

Breeder's equation

$$\begin{align*} &R = h^2i\sigma_P\\ &i = \frac{z}{p}& \end{align*}$$

Breeder's equation

Now, on a time-constant basis to allow comparisons of alternatives:

$$\begin{align*} R & = \frac{i}{L}h^2\sigma_P\\ & = \frac{i_m + i_f}{L_m + L_f}h^2\sigma_P\\ \end{align*}$$

With the $h^2$ and $\sigma_P$ terms possibly constant, we can simply compare the $\frac{i}{L}$ portions under different scenarios of selection and reproduction.

Example: selection for body weight

A poultry breeder is selecting for 56-day body weight in chicken.

• Base population: a random mating population of 154 males and 155 females. Mean value = 1,000g; sd = 50g.

• Selection scheme: 8 males and 48 females with the highest body weight to found the next generation.

• h2: from previous parent-offspring regression and half-sib analysis, $h^2=0.45$.

What is the predicted response to selection in next generation?

$$\begin{align*} & R = \frac{i_m + i_f}{2}h^2\sigma_P\\ \end{align*}$$

i_m <- ifun(p= 8/154) #2.05
i_f <- ifun(p= 48/155) # 1.14
(i_m + i_f)/2*0.45*50

## [1] 35.8358

Example: selection for body weight in chicken

A poultry breeder is selecting for 56-day body weight in chicken.

• Base population: a random mating population of 154 males and 155 females. Mean value = 1,000g; sd = 50g.

• Selection scheme: 8 males and 48 females with the highest body weight to found the next generation.

• h2: from previous parent-offspring regression and half-sib analysis, $h^2=0.45$.

What is the selection differential for the male and female?

$$\begin{align*} &S = i\sigma_P& \end{align*}$$

Sm <- i_m*50 #102
Sf <- i_f*50 # 57

Why response 36g not equal to average value of the $S_m$ and $S_f$?

Example: selection for body weight

• $S_m =102$ and $S_f=57$ are the means of the selected parents
• $R=36$ is the mean of the offspring.

Not equal because the heritability is less than one, which means that the phenotypes of the parents was less than a perfect indictor of their breeding values.

Improvement of response

$$\begin{align*} & R = \frac{i h^2\sigma_P}{L} \\ \end{align*}$$

The form of the breeder's equation allows us to clearly see how to maximize response to selection per unit of time.

1. Reduce the generation interval

This factor is bound by the biology of the organisms, but ways to reduce the generation interval are possible.

• Selection during the off-season in plants

  • winter nursery
  • speed breeding (Watson et al., 2018), up to 6 six generations per year

• Genomic selection

  • Selection without phenotyping

Improvement of response

$$\begin{align*} & R = \frac{i h^2\sigma_P}{L} \\ \end{align*}$$

2. Increase the heritability of the trait

Depends on the genetic architecture, not always under the control of the breeder.

• Maximizing the repeatability of trait evaluation.

• Sound measurement methods and proper experimental design

Improvement of response

$$\begin{align*} & R = \frac{i h^2\sigma_P}{L} \\ \end{align*}$$

3. Increase the selection intensity

• Reduce the number of individuals selected to serve as parents of the next generation

  • Be cautious: increase the inbreeding coefficient and hence loss of alleles through genetic drift.

  • Increase $i$ is likely not the most efficient path! i.e. p from 10% to 5% => i from 1.755 to 2.063.

• Increase the population size from which selections are made

Improvement of response

$$\begin{align*} & R = \frac{i h^2\sigma_P}{L} \\ \end{align*}$$

4. Increase additive genetic variance

$$\begin{align*} & h^2 = \frac{\sigma_A^2}{\sigma^2_P}; & h = \frac{\sigma_A}{\sigma_P} \\ \end{align*}$$

$$\begin{align*} & R = i h\sigma_A/L\\ \end{align*}$$

• The breeding population needs to contain adequate additive genetic variation for the trait of interest.

• If no difference in breeding values exist between individuals within the population, genetic gain through selection is not possible.