Modern interpretation of Johanssen’s results best tackled after mastering some statistics

Intro to statistics

Completed:

Populations and samples

Measures of central tendency

Estimation of genetic and non-genetic causes for phenotypic value

Measures of dispersion and variability

Summarize certain important points we’ve just covered

Starting

Decomposing variability into causal components: Analysis of Variance (ANOVA)

Measures of shared causes: Correlation and Regression

Causal components of variance

Phenotypic variance important in evolution by natural selection

Natural selection acts only on the phenotype

Predicting evolutionary response (breeder’s equation R = h²s )

Requires knowing heritability: ratio of genetic variance to phenotypic variance

Broad-sense heritability is ratio of genotypic variance to phenotypic variance

The proportion of phenotypic variance that is not due to environmental differences

Remember that V_P = V_G + V_E

So, could obtain V_G by subtraction:

V_G = V_P - V_E

Just need to estimate V_E

Class exercise

Design experiment to estimate environmental variance (V_E)

First, generate appropriate equation to estimate environmental variance

Then design experiment to obtain the necessary parameters

Discussion

First, generate appropriate equation to estimate environmental variance

Design necessary experiment

Example using one of Johanssen’s pure breeding lines

Other experiments that estimate V(E)

Nilsson-Ehle (1908)

Is there V(E)?

Zero V(E)

East (1910)

Is there V(E)?

What it would look like if no V(E)

Considerable V(E)

Shortcoming of our technique

Can you discern a shortcoming of the technique we’ve been exploring?

Decomposing variance into causal components

Now use analysis of variance (ANOVA) to partition the phenotypic variance into its causal components

Suppose we have five seeds from each of Johanssen’s 19 pure lines = 95 seeds

Phenotypic values

The observed population mean phenotype is:

Also called the grand mean

Phenotypic variance

Estimate the observed phenotypic variance in a population of 95 individuals by examining the deviation of each individual’s phenotype from the population mean phenotype

This is also called the total mean square = MS_T = (SS_T/df_T)

Phenotypic mean square = total mean square

The phenotypic sums of squares, then, is

Its degrees of freedom are the total experiment size (T = nm) minus 1 (used to estimate the grand mean)

In our experiment d.f. = mn-1 = (19x5)-1 = 95 – 1 = 94

Partitioning phenotypic variance

Phenotypic variance (or mean square) is composed of two causal components

Genotypic variance component

Environmental variance component

To partition phenotypic variance

Need to partition the sums of squares

Partition sums of squares by expanding around the genotype means

Here is the equation for genotypic mean

Simply

Add genotypic mean to one place on right hand side of equation

Subtract genotypic mean to another place on right had side of equation

Because adding one and subtracting one, overall effect is no change in value

What does this trick accomplish?

Produces two kinds of deviation

Within pure line environmental deviations

Deviations of individual seeds from the line mean

Among pure line genotypic deviations

Deviations of each line from the population mean

Now, do the multiplication:

Everything within the bracket can be broken up because

Summation of sums = sum of summations

Right hand side of equation now has three terms

Left hand term is sum of squares of within pure line deviations

Right hand term is sum of squares of among pure line deviations

Center term is crossproduct

Don’t need to worry about what it means because it is the product of two sums of deviations

And, what do deviations sum to?

Further, because the right hand term doesn’t include j, we can sum it over all j

To obtain

Hence, phenotypic sums of squares partitions into two components

Hence,

Sums of squares of environmental deviations

The 1^st term on the right side is the sums of squares of deviations of seeds from their pure line (genotypic) mean

Sums of squares of genotypic deviations

The 2^nd term on the right side is the sums of squares of deviations of line means from the population mean (grand mean)

Partition phenotypic variance by using means squares to estimate the variance components

Estimating environmental variance

Environmental variance can be estimated by the error mean square

In our experiment, d.f. = 95 – 19 = 76

The mean square error is also an estimate of the sampling error,

Estimating genotypic variance is somewhat more complex

Here’s the genotypic mean square

In our experiment, d.f. = 19 – 1 = 18

This value is the observed variance among line means

The observed variance among the genotype means is not the best estimate of the genotypic variance

It is biased by the sampling error of the means

The variance of observed genotypic means is a function the sum of the variance of the true genotypic means, , and their sampling error,

The sampling error of a mean is

To find better estimate of genotypicvariance, go back to genotypic sums of squares

Expected genotypic sums of squares is:

Where

= variance of true genotypic means

and

= sampling error of true genotypic means

The sampling error of a mean is

Hence,

Factor out m :

Divide by degrees of freedom (n – 1) to obtain expected genotypic mean square

Hence, expected mean square is

So, better estimate of true genotypic variance is:

Better estimates of true genotypic and phenotypic variances

Need better estimators of true variances

Expected mean squares point the way to these better estimators:

Broad-sense heritability

We now have estimates of both V_G and V_E

If we assume that G and E are uncorrelated (that is, there is no pattern such that small genotypes are more likely to occur in soils of unusual richness or poorness,

And if we assume that all genotypes interact similarly with all environments (that is, there is no pattern such that some genotypes increase leaf area more than other genotypes in response to improvements in the environment, then:

V_P = V_G + V_E

This leads us to a definition of broad-sense heritability:

The proportion of phenotypic variance due to differences among genotypes in the sample,

Or, the degree to which the phenotype is genetically determined

Source	df	Sums of squares	Observed mean squares	Expected mean squares
Among genotypes	n - 1
Within genotypes	nm - n
Total	nm - 1

Modern interpretation of Johanssen’s results best tackled after mastering some statistics

Intro to statistics

Completed:

Populations and samples

Measures of central tendency

Estimation of genetic and non-genetic causes for phenotypic value

Measures of dispersion and variability

Summarize certain important points we’ve just covered

Starting

Decomposing variability into causal components: Analysis of Variance (ANOVA)

Measures of shared causes: Correlation and Regression

Causal components of variance

Phenotypic variance important in evolution by natural selection

Natural selection acts only on the phenotype

Predicting evolutionary response (breeder’s equation R = h2s )

Requires knowing heritability: ratio of genetic variance to phenotypic variance

Broad-sense heritability is ratio of genotypic variance to phenotypic variance

The proportion of phenotypic variance that is not due to environmental differences

Remember that VP = VG + VE

So, could obtain VG by subtraction:

VG = VP - VE

Just need to estimate VE

Class exercise

Design experiment to estimate environmental variance (VE)

First, generate appropriate equation to estimate environmental variance

Then design experiment to obtain the necessary parameters

Discussion

First, generate appropriate equation to estimate environmental variance

Design necessary experiment

Example using one of Johanssen’s pure breeding lines

Other experiments that estimate V(E)

Nilsson-Ehle (1908)

Is there V(E)?

Zero V(E)

East (1910)

Is there V(E)?

What it would look like if no V(E)

Considerable V(E)

Shortcoming of our technique

Can you discern a shortcoming of the technique we’ve been exploring?

Decomposing variance into causal components

Now use analysis of variance (ANOVA) to partition the phenotypic variance into its causal components

Suppose we have five seeds from each of Johanssen’s 19 pure lines = 95 seeds

Phenotypic values

The observed population mean phenotype is:

Also called the grand mean

Phenotypic variance

Estimate the observed phenotypic variance in a population of 95 individuals by examining the deviation of each individual’s phenotype from the population mean phenotype

This is also called the total mean square = MST = (SST/dfT)

Phenotypic mean square = total mean square

The phenotypic sums of squares, then, is

Its degrees of freedom are the total experiment size (T = nm) minus 1 (used to estimate the grand mean)

In our experiment d.f. = mn-1 = (19x5)-1 = 95 – 1 = 94

Partitioning phenotypic variance

Phenotypic variance (or mean square) is composed of two causal components

Genotypic variance component

Environmental variance component

To partition phenotypic variance

Need to partition the sums of squares

Partition sums of squares by expanding around the genotype means

Here is the equation for genotypic mean

Simply

Add genotypic mean to one place on right hand side of equation

Subtract genotypic mean to another place on right had side of equation

Because adding one and subtracting one, overall effect is no change in value

What does this trick accomplish?

Produces two kinds of deviation

Within pure line environmental deviations

Deviations of individual seeds from the line mean

Among pure line genotypic deviations

Deviations of each line from the population mean

Now, do the multiplication:

Everything within the bracket can be broken up because

Summation of sums = sum of summations

Right hand side of equation now has three terms

Left hand term is sum of squares of within pure line deviations

Right hand term is sum of squares of among pure line deviations

Center term is crossproduct

Don’t need to worry about what it means because it is the product of two sums of deviations

And, what do deviations sum to?

Predicting evolutionary response (breeder’s equation R = h²s )

Remember that V_P = V_G + V_E

So, could obtain V_G by subtraction:

V_G = V_P - V_E

Just need to estimate V_E

Design experiment to estimate environmental variance (V_E)

This is also called the total mean square = MS_T = (SS_T/df_T)

The 1^st term on the right side is the sums of squares of deviations of seeds from their pure line (genotypic) mean

The 2^nd term on the right side is the sums of squares of deviations of line means from the population mean (grand mean)

We now have estimates of both V_G and V_E

V_P = V_G + V_E

In a uniform environment, V_E = 0 and V_P = V_G