Polygenic Inheritance and Heritability.

Plant height in Tobacco is controlled not by a single pair of genes but by a series of genes at multiple loci that each has a small additive affect on the phenotype of the plant. Assume three loci, each of which has two alleles. (A₁,A₂ B₁,B₂ C ₁,C₂). Imagine pure-breeding short plants are all A₂A₂ B₂ B₂C₂ C₂and tall plants are all  A₁A₁  B₁B₁ C₁C₁ and a situation where the height of the plant is determined entirely by the number of upper case alleles regardless of which locus the allele is at. Thus a plant with the genotype A₁A₂ B₁ B₂ C ₁ C₁ is the same height as a plant with genotype A₂A₂ B₁ B₁ C₁ C₁.  This type of model is sometimes referred to as an additive model in that the effects of each gene are small and the the heterozygote at each locus has an intermediate value for a quantitative trait such as height.  Here we might refer to the alleles labeled with a 2 subscript as contributing to short plant height, the alleles with a 1 subscript as contributing to tall height. 

There are 7 possible classes of plant heights depending on the number of alleles.

0,1,2,3,4,5 or 6.

Consider a pure breeding short plant crossed with a pure breeding tall plant. The F1's resulting from this cross are clearly the triple heterozygote:

A₁A₂ B₁ B₂C₁ C₂.

Notice that these plants are going to be intermediate in height between the two parents. This illustrates an important point that many times when you have two parents who differ in phenotype for some characteristic, there is a tendency for the offspring to be intermediate to the parents in phenotype. This phenomenon is sometimes called regression to the mean.

But what happens when these intermediate individuals are bred with each other? To analyze this, assume that the gene pairs are unlinked. This allows us to use independent assortment to predict the results. The expected fraction of offspring in each height class is given by the following expression based on the binomial theorem:

where N is the number of alleles in total(6) and M is the number of tall plant(subscript = 1) alleles in a particular class. One way to interpret this formula is as the number of ways of choosing an individual plant can have M tall plant(subscript = 1) alleles out of N. Sometimes we say N choose M for this.

N for our example is 6. Thus when M is zero there is only one way to get no tall plant(subscript = 1) alleles. But when M is 1 there are 6!/(1!(5!) ) = 6 ways to do this.

Consider when M = 3 as would be the case for the F2's from our cross. Then we have 6! / (3! 3!) = 6*5*4/3*2*1 = 120/6 = 20. All the possible classes out of 64 are shown in the graph here.

So the upshot is this. For the F1's the average phenotypic value for plant height is going to be at the mid point of the heights of the parent plants and if the environment has no effect on the phenotype all of the F1 plants will have that mid point value. But when the F1's are bred together, the resulting F2's still have that mid point value as their average phenotypic value but there is now genetic variation among the offspring since they can now differ in terms of the number of each type of allele.

Heritability.

Often times both genetics and the environment interact to determine the phenotype and geneticists often want to know the degree to which both environment and genetics influence the phenotype in a particular situation. More general models assuming multiple environmental and genetic factors can be constructed to tease out the effects of environment and genetics and the interaction between environment and genetics. These models are called mutlifactorial models.

This can only be done in a statistical sense for populations of individuals and involves the notion of heritability. Heritability in the broad sense H² refers to the fraction of the total variance in a population's phenotypes which is due to genetics. When H² is 1.0 or close to it this says that the variance in a populations phenotype is due entirely to genetics; when H² is close to 0.0 the population's phenotype is due entirely to the variation in the underlying environment but not influenced by genetic variation.

There are several ways to estimate heritability each one makes certain assumptions. One way is to collect the following sorts of population data. The phenotypic variance of a genetically diverse population grown in the full range of natural environments. This gives an estimate of the total phenotypic variance of the population:

Total phenotypic variance = variance due to genetics + the variance due to the environment

Next collect data on a genetically uniform population grown in the full range of environments. Such a population can be obtained from an F1 population from two highly homozygous strains bred together.

This allows us to estimate the variance just due to the environment.

For example this was done for a species of cave fish to estimate the degree to which reduction in eye size was related to genetics vs the environment. Inbred lines from outside the cave and inside the cave were bred to yield genetically uniform F1 to estimate the variance in eye size due to the environments. This turned out to be 0.057. (Genetic back ground assumed constant) These heterozygous F1's were mated and yielded genetically variable offspring and hence an estimate of:

Total phenotypic variance = variance due to genetics + the variance due to the environment

This turned out to be = 0.563

Thus the variance due to genetics = 0.563 - 0.057 = 0.506

And H² = 0.506/0.563 = 0.898

For human quantitative traits, Heritability can often be estimated using monozygotic and dizygotic twins. Since monozygotic twins are genetically identical, any variance in the value of the trait being studied must be due to the environment. Monozygotic twins reared apart then provide an estimate of the variance due to the environment.

In practice they way this is done is by looking at the correlation coefficient ( r ) for the monozygotic twins reared apart. This allows a direct estimate of heritability for the range of environments under consideration. For instance, total ridge count data for the fingerprints of monozygotic twins has a correlation of 0.95. Thus the heritability of this train is 0.95.

It turns out that other types of relationships can be studied in the same sort of way and theoretical correlation coefficients for a multifactorial traits can be worked out. The common ones are given below:

Notice that the fraction by which H² is multiplied to get the correlation coefficient is the fraction of genes the individuals share in common because of descent from a common ancestor. For example, monozygotic twins share all their genes in common, dizygotic twins and full sibs in general share half their genes in common etc.

A distinction is often made between broad sense and narrow sense heritability. Broad sense heritability is the fraction of phenotypic variance due to genetics. Narrow sense heritability is the fraction of phenotypic variance is that due strictly to additive inheritance. For instance if in the plant height example the alleles with subscript 1 were completely dominant then the heterozygotes would be tall. Thus heritability in the broad sense would be smaller because some of the genetic variance is not expressed. The situation as presented involves purely additive inheritance since the heterozygotes are intermediate in phenotype. If the additive model applies then heritability in the broad sense is equal to heritability in the narrow sense.

Heritability estimates must be interpreted carefully. For example a high heritability is often taken to mean that the environment has no influence on a trait. But consider the following hypothetical example for IQ scores of adopted children and their biological and adoptive parents:

In this example the correlation between the biological parent and children's IQ is high, indicating a high heritability and the correlation between the IQ of the adoptive parents and the adoptive child is zero, as one might expect. But observe that the children's IQ is on average closer to that of the adoptive parent score. This sort of situation might arise if adoptive parents are consistently able to provide a better over all environment for the children.

Consider the following example concerning the heritability of learning in two strains of rats that had been selected for high Vs low maze learning ability in normal (for lab rats at least) environments. The low learning ability rat performances in the maze are shown in blue the high maze learning ability rats performances are shown in red. Rats from these genetic strains show clear separation in learning ability in normal environments in which these genetic strains were developed, indicating a strong genetic component to learning. Were one to measure heritability of maze learning for the rats in this environment it would probably be quite high.

But notice what happens when rats from these strains are reared in an enriched environment. The ability of the low maze learning rats increases very significantly where as the learning ability of the fast maze learners increases only slightly. Conversely in an impoverished environment, rats of these strains do equally poorly. Again like the human IQ example, the rat situation emphasizes that heritability estimates are only good for the population and range of environments in which the original studies have been done.

This is an important point for human policy considerations as it relates to education, as sometimes politicians and educational experts sometimes attempt to apply heritability studies in an inappropriate way to justify policy proposals.

pgd 9/26/01 revised 04/20/03