INTRODUCTION TO PLANT BREEDING

AGRONOMY 815 / COURSE NOTES

P. STEPHEN BAENZIGER, 338 Keim Hall, 472-1538

DEPARTMENT OF AGRONOMY / UNIVERSITY OF NEBRASKA

QUANTITATIVE GENETICS


Fehr, Chapt. 6 and 7
Briggs and Knowles, Chapt. 7.


Some traits can be characterized into distinct classes regardless of the environment in which they are observed, e.g., kernel or flower color. For these qualitative characters the phenotype generally gives us a relatively simple expression of the genotype. We can deduce the genotype from genetic or molecular analysis.

1. Each genotype does not have a single phenotypic expression but a norm of reaction -- blurring or phenotypic expression among genotypic classes.

See figures 1 & 2.

2. Many segregating loci whose alleles affect the phenotype being observed.

3. Environmental variation. Two individuals of the same genotype may not have the same phenotype.

Hence quantitatively inherited characters usually are defined as characters under the control of a large number of genes or factors that are similar and relatively small in effect, non-dominant in expression, and acting in a cumulative manner. Examples include height, volume weight, fruit size, winter hardiness, baking or malting quality, potato chipping quality, etc. In other words, the lines show a continuous range in expression making the distinction between phenotypic classes impossible.

The main difference between qualitative and quantitative traits is not necessarily the number of segregating loci but the magnitude of phenotypic differences between genotypes as compared with the individual variation within genotypic classes. This depends not so much on the magnitude of the effect of individual genes as on the relative importance of heredity and environment in producing the final phenotype.

Questions to ask in analysis of quantitative character:

  1. Is the observed variation in the character influenced at all by genetic variation?

  2. If there is genetic variation, what are the norms of reaction of the various genotypes?

  3. How important is genetic variation as a source of total phenotypic variation? (This is a measurement of heredity. Also note breeders only work with genetic modification.)

  4. Are many loci (or only a few) varying with respect to the character? How are they distributed over the genome? (This relates to linkage effects.)

  5. How do the different loci interact with each other in influencing the character? Is there dominance or epistasis?

  6. Is there any non-nuclear inheritance (cytoplasmic effects)?


Basic statistical analysis is essential in the study of quantitative genetics and in its exploitation through plant breeding. Most quantitative characters appear as a normal or continuous curve. However, you occasionally find an interrupted or non-continuous curve:

Important population parameters (characteristics of a population  infinite number of plants measured for the trait of interest., yield).

Mean VarianceStandard deviation
_______________________________________________________________________________________________
s 2 or s2 s or s

Since we are only able to look at samples of the population we can only estimate these parameters.

Sample statistics of interest:

Mean where Xi = ith measurement N = number of the measurements in the sample;
or, if one has a number of individuals falling into particular measurement classes then —

where n = number of individuals in the kth class.

= f1 X1 + f2 X2 + . . . . . + fn Xn =

Variance -- average squared deviations of observations from the mean.

this is not an unbiased estimate of s

or

or

Standard deviation see Fig. 3.

Association or correlation between variables. Because of complex paths of causation, many variables vary together in an approximate way.

e.g., Usual measure of the precision of a relationship between two variables (e.g., X & Y) is the correlation coefficient rXY, calculated from the covariance of X and Y.

Covariance

cov (XY)

= or

Correlation coefficient

An important use of correlation in quantitative genetics is in defining the relationship between family members, e.g., parent — offspring, sib — sib.

Note that in a population: (Two Tailed Distribution)

Mean +/- s is 68% of the population

Mean +/- 2s is 95% of the population

Mean +/- 3s is 99% of the population

This is useful to remember because when you practice selection in cross pollinated crops, you hope to make progress by selecting individuals in the category greater than 1 or 2.

The standard error of the mean is where n is the number that went into the sample. Standard error of the mean can be reduced by 1. increasing the sample size, or 2. reducing the variation in the sample (using controlled environments or more uniform material). Note the importance of the assay.

Two types of gene action are:

  1. Arithmetic — each gene adds or subtracts its contribution to that of the residual genotype. Additive if arithmetically cumulative. Note that the effect of the gene is independent of the other gene at the locus. Additive gene action relates to the average effect of substituting a gene (truly a genic effect)

  2. Multiplicative — each gene or gene substitution multiplies or divides the residual genotypic value by a certain amount. A geometric progression may result. This can be due to dominance or epistatic gene action. Dominant gene action relates to the interaction of genes within a locus (must be two genes to have dominance)Epistatic gene action relates to the interaction between genes at different loci.

It is very unlikely that a simple consistent type of gene action applies to all characters. The effect of any gene depends on its interaction with other genes, its threshold, its dominance relationship, its pleiotropic effect, and its penetrance and its expressivity.

The most important question to be asked about a quantitative trait is whether the observed variation in the character is influenced by genes at all.

Heritability = proportion of the total variance that is due to genetic variance.

This 'broad sense' heritability tells what proportion of the populations variation in phenotype is assignable to variation in genotype. It does not identify the proportion of an individual's phenotype that is heritable.

Estimates of genetic variance are obtained by determining genetic similarity between relatives. Correlation between relatives is evidence for genetic variation only if relatives do not share a common environment more than do non-relatives.

Another important point concerning heritability, H2 is that it is not a fixed characteristic of a trait. H2 of a trait depends upon the population in which it is measured and the set of environments in which the population has developed. i.e., the heritability of a trait is different in each population and each set of environments, and it cannot be extrapolated from one population and set of environments to another.

Further, the separation of variance we observe into genetic and environmental components, sg2 and se2, does not really separate the genetic and environmental causes of variation. Genetic variance depends on the environments to which the population is exposed, and the environmental variance depends upon the frequencies of the genotypes. As a consequence, knowledge of the heritability of a trait does not permit prediction of how the distribution of the trait will change if either genotypic frequencies or environment are changed markedly. Then, why do we bother to consider H2 at all? Mainly because it is a step in understanding the related concept  'narrow sense' heritability.

A finer subdivision of phenotypic variance can provide important information for plant breeders. Genetic and environmental variation can be subdivided in a way that can provide information about gene action and the possibility of shaping the genetic composition of a population.

ADDITIVE AND DOMINANCE VARIANCE

If one of the alleles at a locus codes for a less active gene product or one with no activity at all, and if one unit of gene product is sufficient to allow full physiological activity of the plant, then we expect complete dominance of one allele over the other, as Mendel observed for flower color in peas. If, physiological activity is proportional to the amount of active gene product, we would expect the heterozygote's phenotype to be exactly intermediate between the homozygotes (no dominance). For many quantitative traits, however, neither of these simple cases is the rule. In general, heterozygotes are not exactly intermediate between the two homozygotes but are closer to one of the homozygotes (partial dominance). The complexity of biochemical pathways is such that intermediate activity of primary gene product is not exactly scaled as intermediate phenotype. In some cases the heterozygote may lie outside the phenotypic range of the homozygotes altogether (overdominance).

Let us consider additive and partial dominance in a population. Suppose there are two alleles (a and A) segregating at a locus influencing plant height. In the environments encountered by the population, the mean phenotypes (heights) and frequencies of the three genotypes are:

aa

Aa

AA

Phenotype

10

18

20

Frequency

0.36

0.48

0.16

There is genetic variation in the population; the phenotypic means of the three genotypic classes are different. Some of the variance arises because there is an average effect on phenotype of substituting an allele A for an allele a . i.e., the average height of all plants with A alleles is greater than that of all plants with a alleles. This average effect is easily calculated by counting the a and A alleles and multiplying them by the height of the individuals in which they appear.

.36 of all the individuals are homozygous aa (2 a alleles) with an average height of 10 cm.

.48 of the population is heterozygous (each individual having only one a allele. The average phenotypic measurement of Aa individuals = 18 cm.

Thus, the average effect of all the a alleles is


similarly,

This average difference in effect between A and a alleles of 5.6 cm accounts for some of the variance in phenotype, but not for all, because the heterozygote is not exactly intermediate between the homozygotes. There is some dominance. The total genetic variance associated with this locus can thus be partitioned between the additive genetic variance (sa2) associated with the average effect of substituting A for a, and the dominance variance (sd2) resulting from the partial dominance of A over a in heterozygotes.

i.e., sg2 = sa2 + sd2

The components of variance can be calculated using the definitions of mean and variance given earlier.

fiXi = (.36)(10) + (.48)(18) + (.16)(20) = 15.44 cm.

The total genetic variance that arises from the variation among the mean phenotypes of the three genotypes is

sg2 = S fi(Xi -X)2 = (.36)(10 - 15.44)2 + (.48)(18 - 15.44)2 + (.16)(20 - 15.44)2 = 17.13 cm.

The additive variance is calculated from the squared deviation of the average a effect from the mean, weighted by its frequency, plus the squared deviation of the average A effect from the mean, weighted by its frequency.


sa2 = 2 [ fa(a - X)2 + fA(A - )2 ] = 2 [ (.6)(13.2 - 15.44)2 + .4(18.8 - 15.44)2 ] = 15.05 cm2

sd2 = sg2 - sa2 = 17.13 - 15.05 = 2.08 cm2

This subdivision of genetic variance is useful in the prediction of the effect of selection in breeding.

e.g., Suppose that overdominance is expressed for plant height in the particular population of interest and the phenotypic means and frequencies are as follows:

AA

Aa

aa

Phenotype

10

12

10

Frequency

0.25

0.50

0.25

There is no average difference between a and A alleles because each has an effect of 11 units. i.e., here is no additive genetic variance, although there is dominance variance.

The tallest individuals are heterozygotes. If one attempts to increase plant height in this population by selective breeding, crossing these heterozygotes together will simply reconstitute the original population. Selection will be totally ineffective.

The effectiveness of selection depends on additive genetic variance and not on genetic variance in general.

The total phenotypic variance can now be written as:

sp2 = sg2 + se2 = sa2 + sd2 + se2.

And, heritability h2, heritability in the narrow sense, is defined as follows:

It is this heritability, not to be confused with H2, that is useful in determining whether a program in selective breeding will succeed in changing the population.

Remember:

"The concept of additive variance does not carry with it the assumption of additive gene action: and the existence of additive variance is not an indication that any of the genes act additively (i.e., show neither dominance, nor epistasis). No assumption is made about the mode of action of the genes concerned. Additive variance can arise from genes with any degree of dominance or epistasis, and only if we find that all the genotypic variance is additive can we conclude that the genes show neither dominance nor epistasis."

Falconer, p 118.

Also self-pollinated crop breeders are generally most interested in additive genetic variance because it can be fixed in homozygous lines. Also as only one gene comes from each parent. Hence the ability to understand the additive genetic variance and the average effect of a gene explains to how a parent relates to its progeny. Broad sense heritability is useful as an upper limit estimate of narrow sense heritability.

ESTIMATING GENETIC COMPONENTS OF VARIANCE

Genetic components of variance are estimated from covariance between relatives. Epistatic contributions to variance are contained to some degree in all covariances between relatives but this is difficult to estimate and is usually ignored.

CAN QUANTITATIVE INHERITANCE BE EXPLAINED IN WITH MENDELIAN GENES?

A continuous curve can be obtained by expanding the binomial:

(a + b)2 n where n = number of segregating loci. Consider 6 segregating loci then:

(a + b)12 = a12 b0 +12a11 b1 +66a10 b2+220a9 b3+
495a8 b4+792a7 b5+924a6b6+792a5b7+495a4b8+
220a3b9+66a2b10+12a1b11+a0b12

The coefficients are 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1 which can be expressed as a % of the total as: 0, 0.2, 1.7, 5.4, 12.2, 19.3, 22.4, 19.3, 12.2, 5.4, 1.7, 0.2, 0.

The same coefficients for a normal curve are: 0, 0.3, 1.6, 5.2, 11.8, 19.5, 22.9, 19.5, 11.8, 5.2, 1.6, 0.3, 0.

s a2 = 2pq [a + d(q-p)]2s d2 = (2pqd)2

General consequences of plant size inheritance (East):

  1. The F1 in quantitative inheritance studies will generally be intermediate between parents (lack of dominance). It will be as uniform as the parents (assumes parents are homozygous). Sometimes the F1 is less affected by the environment, hence is more uniform than the parents.

  2. The F2 population will be more variable than the F1 but their means should be similar.

  3. The F2 consists of a series of grades over the range of the parents and may even exceed them by transgressive segregation (putting all the favorable alleles together — complimentary gene action).

  4. The relative number of grades in the F2 follows the frequency of 2n + 1 where n = the number of loci that the parents differ. Assumes each gene has similar effects. 3 loci segregating: 2(3) + 1 = 7.
  5. The number of individuals in the different grades may be gotten by the expansion of the binomial coefficients. When no dominance occurs, a normal curve results. With dominance, a skewed curve results.

  6. The large portion of the F2 will be intermediate and will resemble the F1 .

An example of quantitative genetics being explained in a Mendelian fashion is as follows. Assume you cross the following genotypes:

aabbCC (30 inches tall)xAABBcc (40 inches tall).

Also assume the effect of each allele is the same for plant height (for example: AAbbcc, aaBBcc, AaBbcc, AabbCc, aabCC all are 30 inches tall). What is the effect of each allele?

Answer: aabbCC and AABBcc differ by a total of two "effective alleles" and by 10 inches. Each effective allele adds 5 inches to the height.

What is the height of the F1: AaBbCc? It has three effective alleles for 15 inches of height. Also, aabbcc would be 10 inches shorter than aabbCC. Hence AaBbCc would be 20 + 15 = 35 inches tall.

How many classes are there in the F2 and what are their proportions?

The number of classes is 2n + 1 where n is the number of segregating loci. 2(3) + 1 = 7 classes.

(a + b)2 n = (a + b)6 = 1a6 b0 + 6a5 b1 + 15a4 b2 + 20a3 b3 + 15a2 b4
+ 6a1 b5 + 1b6

Hence, there would be (1) 20-inch plant, (6) 25-inch plants, (15) 30-inch plants, (20) 35-inch plants, (15) 40-inch plants, (6) 45-inch plants, and (1) 50-inch plant.

Can you work the problem if the following cross had been made:

aabbCCxaaBBCC.