**Variance Components**

Consider the simple, single-locus model (below) with alleles *b*^{1} and *b*^{2}.

The midparent value *m* = (1/2)(*b*^{1}*b*^{1} + *b*^{2}*b*^{2}). If the heterozygote does not have a phenotypic value equal to *m*, some degree of dominance (*d*) exists. If no dominance exists, then the alleles are completely additive. However, quantitative traits are governed by many loci and it might be possible that genotype *b*^{1}*b*^{2} is dominant in a positive direction whereas genotype *c*^{1}*c*^{2} is dominant in a negative direction, so that they cancel each other, giving the illusion of additivity. Dominance of all types can be estimated from the variances of F_{2} and backcross generations. All of the phenotypic variance within pure lines *b*^{1}*b*^{1} and *b*^{2}*b*^{2}, as well as in their genetically uniform F_{1} (*b*^{1}*b*^{2}), is environmental. Hence, the phenotypic variances of each pure parental line (*V _{P1}* and

*V*) as well as that of the F

_{P2}_{1}(

*V*) serve to estimate the environmental variance (

_{F1}*V*). The F

_{E}_{2}segregates (1/4)

*b*

^{1}

*b*

^{1}: (1/2)

*b*

^{1}

*b*

^{2}: (1/4)

*b*

^{2}

*b*

^{2}. If each genotype departs from the midparent value as shown in the above model, then the average phenotypic value of

*F*

_{2}should be (1/4)(–

*a*) + 1/2 (+

*d*) + 1/4 (+

*a*) = (1/2)

*d*. The contribution that each genotype makes to the total is its squared deviation from the mean (

*m*) multiplied by its frequency . Therefore, the total F

_{2}variance (all genetic in this model) is the mean of squared deviations from the mean:

If we let *a*^{2} = *A*; *d*^{2} = *D*, and *E* = environmental component, then the total F_{2} phenotypic variance (*V _{F2}*) = (1/2)

*A*+ (1/4)

*D+E*, representing the additive genetic variance (

*V*) + the dominance genetic variance (

_{A}*V*) + the environmental variance (

_{D}*V*), respectively. Likewise it can be shown that

_{E}*V*(the variance of backcross progeny F

_{B1}_{1}× P

_{1}) or

*V*(the variance of backcross progeny F

_{B2}_{1}× P

_{1}), = (1/4)

*A*+ (1/4)

*D+E*, and V

_{B1}+ V

_{B2}= (1/2)

*A*+ (1/2)

*D+2E*. The degree of dominance is expressed as

Heritability can be easily calculated from these variance components. The same is true of variance components derived from studies of identical (monozygotic) vs. nonidentical (fraternal, dizygotic) twins. If twins reared together tend to be treated more alike than unrelated individuals, the heritabilities will be overestimated. This problem, and the fact that the environmental variance of fraternal twins tends to be greater than for identical twins, can be largely circumvented by studying twins that have been reared apart.

**Genetic Similarity of Relatives**

If offspring phenotypes were always exactly intermediate between the parental values regardless of the environment, then such traits would have a narrow heritability of 1.0. On the other hand, if parental phenotypes (or phenotypes of other close relatives) could not be used to predict (with any degree of accuracy) the phenotypes of offspring (or other relatives), then such traits must have very low (or zero) heritabilities.

**Regression Analysis**

The **regression coefficient** (*b*) is an expression of how much (on the average) one variable (*Y*) may be expected to change per unit change in some other variable (*X*).

EXAMPLE 8.7If for every egg laid by a group of hens (X) the average production by their respective female progeny (Y) is 0.2, then the regression line ofYonXwould have a slope (b) of 0.2 (symbol Δ = increment of change).

The regression line of *Y* on *X* has the formula

where *a* is the "*Y* intercept" (the point where the regression line intersects the *Y* axis), and and are the respective mean values. The regression line also goes through the point (, ) and establishing these two points allows the regression line to be drawn. Any *X* value can then be used to predict the corresponding *Y* value. Let = estimate of *Y* from *X*; then

Since daughters receive only a sample half of their genes from each parent, the daughter-dam regression estimates only one-half of the narrow heritability of a trait (e.g., egg production in chickens). If the variances in the two populations are equal (*S _{x} = S_{y}*), then

Similarly, the regression of offspring on the average of their parents (midparent) is also an estimate of heritability

Full sibs (having the same parents) are expected to share 50% of their genes in common; half-sibs share 25% of their genes. Therefore,

If the variances of the two populations are unequal, the data can be converted to standardized variables (as discussed later in this chapter) and the resulting regression coefficients equated to heritabilities as described above.

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