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Hello. In this

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example, we will

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calculate the estimation of coefficient

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of heritability using analysis of

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variance on performance values in

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related individuals.

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In our case, we use groups

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of half sibs by sires.

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40 progeny were chosen at random.

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Each sire mated to 8 dams, with each

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mating producing one male.

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Five sire families were chosen at

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random and progeny 8 week

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body weight in grams

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obtained.

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Genetic variances and heritability

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are to be estimated. We use

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analysis of variances.

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In the first table we can see 40

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individuals values of performance.

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In this statistical analysis we use

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simple linear model with one 

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effect of fathers.

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Our linear model is "y" which is the

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performance rate of the jth offspring

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and this "y" is equal to means

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plus the effect of i-th 

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father

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plus the residue of

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ei.

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These are random effects that we are

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unable detect and include in the

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calculation.

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To simplify the manual calculation, we

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don't have to perform all the basic

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operations. We have them pre-calculated

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in the second table, where the sum

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groups, sums of squares and

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squares of sums are shown.

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Also the total sums.

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We then use these values in the

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equations to calculate the sum of squares

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of deviations from the mean.

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For calculation we use program

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R and Rstudio which

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you can see on your left.

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We have written the impute values as

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objects with name.

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A large Y squared means the

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square of the total sum.

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The next value is the square of the sum

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for the group divided by the number of

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individuals in the group and summed

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over all groups.

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The third value is the sum of squares of

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the individual values for the

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entire data set.

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P is number of fathers,

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5, N is number of

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individuals, 40,

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and n0 is weighted

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number of offspring per sire.

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These objects must be loaded into the

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computer's memory by selecting them with

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the mouse and running the script by

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pressing the Control and

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Enter key combination.

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We must now calculate the sum of squares

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of the deviation from the mean according

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to the linear ANOVA equation

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first. We calculate the sum of squares

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of the deviations from the mean between

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groups of half sibs. It is between

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the fathers. We name them

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SSA.

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The result is displayed by pressing the

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Control key and Enter.

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Next, we compute the sum of squares

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of the deviations from the mean

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within the groups of half sibs

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SSE.

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After creating the script according to

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the formula, we mark it and run

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it. Again we can see the

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results below.

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The next step is to count the mean of

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square, which are the

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variances,

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and again we calculate the

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mean of square MSa

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between groups and mean of

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square residual MSe

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within groups.

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We calculate the mean of square between

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groups of fathers MSa

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by dividing SSa by the degrees of

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freedom. In this case

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is d the number of fathers

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minus 1.

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We obtain the residual mean of

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square MSe by dividing

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SSe by the degrees of freedom.

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It is the number of offspring minus

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the number of sires.

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And this is the result of statistical

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analysis of variance.

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The genetic part follows where the last

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column of the ANOVA table describes

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the composition of the mean of square.

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The genetic variance contain it in mean

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of square of sires MSa

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and MSe residual mean

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of square directly equal to the

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environmental variance.

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We can, therefore, estimate the

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genetic variance as the difference

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between variance between

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sires, MSA, and

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variance residual,

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MSE. And this

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difference is divided by n0 -

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weighted number of offspring per sire.

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So our genetic variation by

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sire is.

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245.7

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So, let's write down that the

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environmental variation is directly

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equal to the MSE, residual

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mean of square.

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We have now calculated both

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components of variation,

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genetic and environmental.

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From these both variations we have

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to calculate the so-called intraclass

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correlation coefficient.

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We will mark this with an R

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or rho.

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This coefficient is equal to genetic

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variation divided by the total

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phenotype phenotypic variance, which is

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the genetic plus environmental

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variance.

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This equation is very reminiscent of the

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heritability, because heritability is

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defined as proportion of genetic variance

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to total phenotypic variances.

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Why is that not heritability?

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It is because we are calculated

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based on groups of half sibs.

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The value of heritability, h^2,

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is straight four times this

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intraclass correlation coefficient. The

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reason is simple, because the genetic

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similarity between half sibs

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are 25%, 1/4.

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We also have to calculate the value of

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the standard error of estimation of the coefficient of

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heritability. The formula is again

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in the Word document.

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As you can see, the

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standard error of estimation of

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heritability

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is very high -

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0.556.

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In order for us in that calculated value

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of heritability we get a little bit of

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faith, so the standard error would have

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to be less than 0.05.

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In our case, far from it.

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It's because it's only a model example.

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We have got a little offsprings, few

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groups of half-sibs,

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which is why the standard

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error is so high.

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If we use a statistical software like R,

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we don't have to calculate ANOVA

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manually, but we can use the function.

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But first, you have to load the data.

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We have the database prepared in Excel

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spreadsheet that is available to you,

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so you can... It is

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very simple. In RStudio

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load the data. So we

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are gonna put FILE up here.

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We will put Import dataset and select the

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menu From Excel. We

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will find our Excel file.

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As you can see we have already got the

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data. The table is loaded nicely.

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I have a father, sire here in

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the ABCD form and in the second

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column are the values of the

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offspring. We call this data

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data1. We will

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just put import.

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We write object ANOVA1

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and we use function lm, linear

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model, which is the basic function in

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R. We have to define

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model equation. "y"

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is equal to one effect,

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effect of sire. It's

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not used here a call to but the

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so-called tilde or the waveform.

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Write Alt+1 on alphanumeric

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keyboard. To view the

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result, let's write down

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ANOVA and in parenthesis

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ANOVA1. When I

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tag it and run it, we see the

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results in couple of steps,

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actually in two lines. We have

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the resulting table of analysis of

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variance. Notice that here are

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the degrees of freedom, paternal

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or between groups, and the residual.

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Then there is the sum of squares of

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deviations from the mean.

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The last third column is crucial for us.

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There are mean of squares

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between and within groups.

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You can see that this is absolutely the

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same as we can calculate it manuály before.

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This is where the analysis of variance

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ends and then the genetic analysis comes

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in. So we

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should calculate genetic variances,

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environmental variances and then

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coefficient of correlation and from this

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coefficient, coefficient of heritability.

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At the end we will show a second analysis

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where we use the REML method. Here we

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have to load or install and load

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special package lme4.

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You install the package using the package

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folder.

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I have already installed it, so I just

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activate it by typing library

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and the package name.

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Here we will use the lmr

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function that is in the package

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lmf4 and

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again we have to define the simple

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equation. So again

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"y" tilde

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1 plus and in

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brackets it will be

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one, perpendicular line and

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then we will write sire.

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Let's also write that the data is Data1

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to get the result back

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displayed, so we will write a summary of

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our REML1 calculation.

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And if we work it

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and put Ctrl key and Enter, we will

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see the overall table for this analysis,

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where we can notice that these two lines

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are important to us when, as you can

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see, us directly this

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calculation REML calculated directly the

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genetic variation by sire.

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There is also environmental

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variation. We then

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process them in the same way. It is we

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calculate the coefficient r and

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from it the heritability.

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And thank you for your attention.