Quantitative Genetics

Abstract

The study of several economically important traits cannot be addressed by the classic or Mendelian genetics since several genes are involved at the same time, every one with a small individual effect. These traits are referred to as polygenic, and they are characterized by continuous variation and high environmental influence. Because it is not possible to make a natural separation by phenotype in a segregating generation, it is necessary to seek support in statistics for the study of these polygenic traits. The phenotype of an individual depends on the genotype it possesses and on the influence of the environment. In a segregating generation, it is noted that the phenotypic variance is a function of genotypic variance and environmental variance. Phenotypic variance is a measure of the variation that exists at the phenotypic level within a population, the genotypic variance is a measure of variation due to the different genotypes occurring in a given population, and the environmental variance is a measure of the phenotypic variation due to environmental influence. Only genotypic variance is heritable, which is why this term is important in genetic improvement. To facilitate the comparison between populations or between traits, and because the absolute value of the variances is not indicative of the level of variation, the concept of heritability is used, which is the fraction of the phenotypic variance explained by the genotypic variance. By performing repeated agronomic trials in time and space, the phenotypic variance is not only a function of the genotypic variance and of environmental variance, it is also explained by the variance due to the genotype-by-environment interaction (G × A), which is the differential behavior of genotypes in different environments.

Solved Problems

• 1. Suppose that you have been commissioned to evaluate the percentage of sucrose in a sugarcane clone by taking 100 samples in one hectare. The results tell you that the 100 samples differed significantly in the value obtained. To what can you attribute this situation, and is it usable in a program of genetic improvement of sugar cane?

Sugar cane is an asexual reproductive species, in which a type of cultivar called a clone is used. All the individuals of a clone, and in this case, all of the samples taken in the hectare, are genetically equal. Because of this, and knowing that the phenotype is a function of the genotype and environmental influence, it is it can be stated that the observed variations are due to the environment. There will exist in the hectare differences in the soil, or in the humidity, or in the topography, or any other environmental causes that have a differential influence on the individuals of which samples were taken, thus expressing a different phenotype even though they had the same genotype. This variation is not exploitable in a genetic improvement program since environmental variation is not heritable.

• 2. Two homozygous lines of bean were crossed, obtaining the F1, which, being self-pollinated allowed obtaining the F2 generation. By establishing a trial, the following data are obtained from 10 plants of each genotype

Number of pods/plant
	N °of plant
	1	2	3	4	5	6	7	8	9	10
Parent 1	16	17	15	18	19	17	14	16	18	19
Parent 2	11	12	10	10	12	15	16	17	12	11
F2	16	10	11	18	9	19	21	15	12	14

Number of seed/pod
	N ° of plant
	1	2	3	4	5	6	7	8	9	10
Parent 1	4	3	3	3	5	6	4	3	3	4
Parent 2	1	2	2	1	1	2	2	2	2	1
F2	3	3	3	4	4	3	5	2	2	2

• (a) Determine the components of phenotypic variation in each characteristic.

${\sigma}_{\mathrm{P}}^2={\sigma}_{\mathrm{G}}^2+{\sigma}_{\mathrm{E}}^2$. The parents are homozygous lines, so it is assumed that the entire variation among individuals from the same parent is an estimate of the environmental variance. Calculating the variance in both parents will result in two environmental variance estimates, which are averaged. In the F2, several genotypes segregate, and additionally, they are influenced by different environmental conditions, so that the variance in this generation is the sum of the environmental variance and genetic variance, i.e., it is an estimate of the phenotypic variance. Having the phenotypic variance and the environmental variance, genotypic variance can be obtained.

Number of pods/plant

$$ {\displaystyle \begin{array}{l}{\sigma}_{\mathrm{E}}^2=\left({\sigma}_{\mathrm{PARENT}\ 1}^2+{\sigma}_{\mathrm{PARENT}\ 2}^2\right)/2\kern0.78em {\sigma}_{\mathrm{PARENT}\ 1}^2=2.77\kern0.78em {\sigma}_{\mathrm{PARENT}\ 2}^2=6.27\\ {}{\sigma}_{\mathrm{E}}^2=\left(2.77+6.27\right)/2=4.52\end{array}} $$

$$ {\displaystyle \begin{array}{l}{\sigma}_{\mathrm{P}}^2={\sigma}_{\mathrm{F}2}^2\kern0.78em {\sigma}_{\mathrm{F}2}^2=16.28\kern0.78em {\sigma}_{\mathrm{P}}^2={\sigma}_{\mathrm{G}}^2+{\sigma}_{\mathrm{E}}^2\kern0.78em {\sigma}_{\mathrm{G}}^2={\sigma}_{\mathrm{P}}^2-{\sigma}_{\mathrm{E}}^2\\ {}\kern14.375em =16.28-4.52\\ {}\kern14.375em =11.76\ \end{array}} $$

Number of seed/pod

$$ {\displaystyle \begin{array}{l}{\sigma}_{\mathrm{E}}^2=\left({\sigma}_{\mathrm{PARENT}\ 1}^2+{\sigma}_{\mathrm{PARENT}\ 2}^2\right)/2\kern0.78em {\sigma}_{\mathrm{PARENT}\ 1}^2=1.07\kern0.78em {\sigma}_{\mathrm{PARENT}\ 1}^2=0.27\\ {}{\sigma}_{\mathrm{E}}^2=\left(1.07+0.27\right)/2=0.67\end{array}} $$

$$ {\displaystyle \begin{array}{l}{\sigma}_{\mathrm{P}}^2={\sigma}_{\mathrm{F}2}^2\kern1.625em {\sigma}_{\mathrm{F}2}^2=0.90\kern1.625em {\sigma}_{\mathrm{P}}^2={\sigma}_{\mathrm{G}}^2+{\sigma}_{\mathrm{E}}^2\kern1.625em {\sigma}_{\mathrm{G}}^2={\sigma}_{\mathrm{P}}^2-{\sigma}_{\mathrm{E}}^2\\ {}\kern16.25em =0.90-0.67\\ {}\kern16.25em =0.23\end{array}} $$

• (b) On the basis of the phenotypic variance components, which trait is more feasible to be genetically improved? Explain

Absolute values of the variances calculated are not indicative of the suitability or not to genetically improve a population. Even less so in this case where the units are different. In order to be able to make a comparison, traits would have the same unit of measure, with heritability in the broad sense for both traits being a parameter that allows comparison between characteristics and even between populations.

Number of pods per plant

$$ {h}^2=\left({\sigma}_{\mathrm{G}}^2/{\sigma}_{\mathrm{P}}^2\right)\times 100\kern0.78em {h}^2=\left(11.76/16.28\right)\times 100\kern0.78em {h}^2=72.24 $$

Number of seeds per pod

$$ {h}^2=\left({\sigma}_{\mathrm{G}}^2/{\sigma}_{\mathrm{P}}^2\right)\times 100\kern0.78em {h}^2=\left(0.23/0.90\right)\times 100\kern0.78em {h}^2=25.56 $$

The feasibility of genetically improving a population is given by the amount of genetic variation in the population. The number of pods per plant has a heritability much higher than the number of seeds per pod, which indicates that a greater proportion of the phenotypic variation observed is a consequence of the genotypic variation. The number of pods per plant has a better perspective of being genetically improved than the number of seeds per pod.

• 3. When studying the yield in populations A, B, and C of maize, we obtain in A a phenotypic variance of 1850 and a genotypic variance of 1678; in B we obtain a phenotypic variance of 3456 and an environmental variance of 3000; and in C a genotypic variance of 780 and an environmental variance of 20. Indicate in which of the populations it is more feasible to carry out genetic improvement. Justify your answer.

In order to compare populations, heritability in a broad sense must be calculated in the following way:

Population A

$$ {\sigma}_{\mathrm{P}}^2=1850\kern0.78em {\sigma}_{\mathrm{G}}^2=1678\kern0.78em {h}^2=\left(1678/1850\right)\times 100\kern0.78em {h}^2=91 $$

Population B

$$ {\displaystyle \begin{array}{l}{\sigma}_{\mathrm{P}}^2=3456\kern0.78em {\sigma}_{\mathrm{E}}^2=3000\kern0.78em {\sigma}_{\mathrm{G}}^2=3456-3000=456\\ {}{h}^2=\left(456/3456\right)\times 100\kern0.78em {h}^2=13.19\end{array}} $$

Population C

$$ {\sigma}_{\mathrm{G}}^2=780\kern0.78em {\sigma}_{\mathrm{E}}^2=20\kern0.78em {\sigma}_{\mathrm{P}}^2=780+20=800\kern0.78em {h}^2=\left(780/800\right)\times 100\kern0.78em {h}^2=97.50 $$

Yield in population C has the highest genotypic variability, which is indicative of the greater value it has for genetic improvement. Nearly all of the variation observed in the phenotype “yield” in the population C is due to genotypic variation, which increases the likelihood of finding agronomically valuable genotypes.

• 4. The following are three situations in which statistical significance is indicated of three analyses of variance obtained from three trials replicated in time and space (ns stands for non-significant differences, * stands for significant differences):

Variation source	Situation 1	Situation 2	Situation 3
Genotype	*	*	*
Genotype × Location	ns	*	*
Genotype × Year	ns	ns	*
Genotype × Year × Location	ns	ns	*

Which would be your recommendation for each situation?

Situation 1

The interactions are not significant, i.e., when comparing the behavior of the genotypes between them was stable: the best genotypes were good (when compared with the others) in all the environments, those genotypes that were deficient had problems in all environments, maintaining the relative order and the differences between them across locations and years of evaluation. In this case the recommendation is obvious: if one genotype was the best in all locations and in all years, this should be the genotype to be recommended to all farmers in the geographical region where the evaluation took place.

Situation 2

Since there is at least one significant interaction, the conclusions should be based on interaction, without considering what happened with the individual factors. In this case, the Genotype x Location interaction was significant. This suggests that at least in the comparison of a couple of lines, the behavior was differential, that is, perhaps the best line in one environment was not the best line in another environment (crossover interaction), or the difference in performance between a pair of genotypes was not equal in at least one pair of environments but the performance order among genotype was kept through the different environments (non-crossover interaction). The first thing to check is whether the interaction was either non-crossover or crossover. If it was a non-crossover interaction, it is necessary to identify which genotype was best across environments and recommend it. If the interaction was due to specific adaptation of a specific genotype(s) to a specific environment(s) a crossover interaction is taking place, so the recommendation should be made by location: for each location one genotype was the best one.

Situation 3

The most complex interaction is significant. This means that the differential behavior of the genotypes was due to changes in the environmental conditions both in time and space. In order to make recommendations, the three following factors should be considered at the same time: genotypes, locations, and years. It is necessary to consider by means of the records of the trials which conditions were different from year to year. If the years were different due to unpredictable conditions (rainfall, pest attacks, pathogens), genotypes that withstood those conditions better can be identified, and could serve as sources of genes for future breeding programs at sites where the aforementioned conditions will very frequently occur; but it does not mean recommendations can be made for such a genotype because the conditions made different that year are unpredictable. If the conditions are predictable, for example, particular management (fertilization, irrigation, in plant production, nutrition plan in animal), it is possible to recommend a specific genotype in a given location under a given management.