Heritability

In genetics, heritability is the amount of phenotypic variation in a population that is attributable to individual genetic differences. Heritability, in a broad, general sense, is the ratio of variation due to differences among genotypes to the total phenotypic variation for a character or trait in a population. It is expressed as

The range of values for heritability estimates is 0 to 1. If H = 1, then we are able to say that all variation in a population is due to differences or variation among genotypes (i.e., there is no environmentally caused variation). On the other hand, if H = 0, there is no genetic variation. In this case, all variation in the population is from differences in environments during the life experience of individuals. In other words, we can say that all individuals are the same with regard to the effect of their genes on phenotypic variance.

The following example, taken from the volume by Bodmer and Cavalli-Sforza (1976), demonstrates how the concept of heritability can be used to measure the relative contribution of genetic and environmental factors in patterning phenotypic variation.

[Page 491]In the example, we have two genotypes, G1 and G2, and two environments, E1 and E2. Table 1 shows four individuals, two of each genotype, distributed in the two environments, and their phenotype scores for some measurable character or trait.

If we graph these data, they look as shown in Figure 1.

Figure 1 Phenotype Scores Plotted by Genotype and Environment

Source: Adapted from Bodmer and Cavalli-Sforza (1976).

One should note that there is variation, that is, there are differences, among the four individuals on their phenotype scores. Note also that the scores of G1 are higher on average and E1 seems to produce higher phenotype scores, on average, than E2. If we calculate a mean for all four individuals, we get 52/4 = 13 = the mean phenotype score. How would we now express the variation among the four individuals?

We calculate a measure of variation called the variance. The steps are as follows:

• 1.
  First, calculate the deviations of each of the individual scores from the mean.
• 2.
  Then square these deviations.
• 3.
  Then add up these squared deviations to get the sum of squares or SS, which is SS = (16 − 13)2 + (14 − 13)2 + (12 − 13)2 + (10 − 13)2 = 9 + 1 + 1 + 9 = 20.

This SS value is not the variance estimate, but if we divide by 4 − 1 or 3, we have the actual variance, that is, the value for the total variation among phenotypes.

Once we have calculated total phenotypic variance, we can then estimate how this variance partitions into genetic and environmental components. Only the genetic component will be shown here. We obtain the estimate of the genetic variance component by eliminating the environmental source of variation, that is, placing genotypes in either E1 or in E2. Table 2 shows the four individuals placed in environment E1.

If we calculate a mean for the phenotype scores, the value is 14. The SS = 16. These squared deviations are totally the result of variation among genotypes. We can label this second SS as SSg to denote the genetic SS. It should be noted that if we had used E2, we would have obtained the same value of 16 for the SSg. Now, using our SS values, we can calculate the actual broad heritability estimate. It would be Vg=Vt or 16/3 divided by 20/3 = 0.80. In this constructed example, we compute that 80% of the total variance is the result of variance among the genotypes. An important additional point is that this heritability would be valid only under the set of conditions we have shown.

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