Lecture 47: Quantitative Genetics II. The Resemblance between Relatives
(Current Version: 1 December 1999)
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Fisher's Decomposition of the Genetic Value
R. A. Fisher (1918) showed that the genetic value G can be further decomposed as
G = u +A + D
where
- u = Population mean
- A = Additive genetic value, also called the Breeding value
- D = Dominance value
- There are additional terms if epistasis is present (for example, if the genotypic value of a two locus genotype AaBb does not equal the average value of Aa genotypes + the average value of Bb genotypes).
Hence,
z = G + E = u + (A+D) + E
Fisher's great insight :
Information for relatives allows us to estimate
Var(A), Var(D). Using these estimates, we can predict the resemblance between different sets of relatives.
Genetic Covariance between relatives
Consider the covariance between the genotypic values of two relatives (R1 and R2). By construction A and D are uncorrelated.. Hence,
Cov(GR1, GR2)
= Cov(AR1+DR1,
AR2+ DR2)
= Cov(AR1, AR2) +
Cov(DR1, DR2)
If two relatives share only one allele ibd, then
- Cov(AR1, AR2) = Var(A)/2
- Cov(DR1, DR2) = 0
If two relatives share both alleles ibd, then
- Cov(AR1, AR2) = Var(A)
- Cov(DR1, DR2) =Var(D)
Just what do A and D represent?
Additive effects
Consider the genotypic value for a given locus with (potentially) many alleles,
B1, ..., Bk.
(In a random-mating population), The additive effect of an allele Bi is simply the mean genotypic value for an individual carrying a copy of allele Bi.
Hence, the predicted genotypic value of Bi Bj is
Predicted[Gij]= u + ai + aj
Of course, the above are predicted values. The difference between the predicted and actual values for the genotypic value at each locus is defined as the dominance deviation
D ij = Gij - Predicted[Gij]
= G ij - ( u + ai + aj)
Thus to obtain A and D, we simply sum over all loci.
Breeding Values
A is called the breeding value (BV).
- The average breeding value of a random individual is zero
- The expected offspring mean for two parents is the population mean plus the average of the parental breeding values.
- If Fred's BV is 20 and Mary's is -15, then their offspring are expected to have an average value that is (20 - 15)/2 = 2.5 units above the population mean
- The expected offspring mean for one parent is the population mean plus the half the parental breeding value
- Offspring with Fred as their father are expected to be 20/2 = 10 units above the population mean
- Offspring with Mary as their mother are expected to be (-15)/2 = 7.5 units below the population mean
The resemblance between relatives
Since z = u + A + D + E, where u is a constant and A, D, and E are uncorrelated random variables, using the rules of covarinces (Lecture 46) the covariance between two relatives is given by:
Cov(zR1, zR2) =
Cov(AR1, AR2) +
Cov(DR1, DR2) +
Cov(ER1, ER2)
- Covariance of Parent (P) and Offspring (O)
- A parent and its offspring share exactly one allele ibd. Hence
- Cov(P,O) = (1/2) Var(A) + Cov ( EP, EO)
- The environmental covariance typically only occurs with maternal effects, i.e. P = mother.
- Covariance of two Half-sibs (typically the same father)
- Probability that two half-sibs share one allele ibd = 1/2. Else, they share no alleles. Hence
- Cov(HS1, HS2) = (1/4) Var(A)
- Covariance of two full-sibs
- With probabilities 1/4, 1/2, and 1/4, the sibs share 0, 1, and 2 alleles ibd
- Cov(FS1, FS2) = (1/2) Var(A) + (1/4) Var(D)
+ Cov ( E FS1 , E FS2 )
- Again, shared maternal effects can appear as environmental covariances
- dizygotic (i.e., nonidentical) twins are an example of full-sibs,
but they may share greater environmental covariances because the twins have experienced
the same maternal environment
- Covariance of two identical ( monozygotic twins)
- Here, all alleles are ibd, giving:
- Cov(MZ1, MZ2) = Var(A) + Var(D)
+ Cov ( E MZ1 , EMZ2 )
Example: Estimating additive and dominance variances
Suppose :
- the population variance in a trait is 100
- The covariance between (paternal) half-sibs is 10
- The covariance between (maternal) half-sibs is 10
- The covariance between full sibs is 25
What are Var(A), Var(D)? Are there any shared environmental effects?
-
Var(A) = 4*Cov(half-sibs) = 4*10 = 40
- Since maternal and paternal half-sibs the same, no maternal effect
- Cov(Full sibs) = 25 = Var(A)/2 + Var(D)/4 or
- Var(D) = 4* [ Cov(Full sibs)-Var(A)/2 ] = 4*(25-20) = 20
Thus, 80 percent of the total variation is genetic, 40 percent of the total variation is due to additive effects, 40 percent due to dominant effects.
The concept of heritability
- The total fraction of phenotypic variation accounted for by genetic effects
- Some examples:

Estimating heritability: parent-offspring regressions
- Idea: if there is some genetic basis to the character, then offspring should resemble their parents
- Galton: one measure of this is to regress parents on offspring
- Fisher (1918) connected these methods of resemblance between relatives from the biometricians with mendelian models of genetics

Galton' s data on human height
- The expected slope of the best linear fit of the value of a single parent on the mean value of their offspring is h2/2
offspring mean = pop mean + ( h2/2 ) ( parent value - pop mean )
- The expected slope of the midparental value (average of the two parents) on the mean value of their offspring is h2.
offspring mean = pop mean + ( h2 ) ( midparent value - pop mean )
Example
Suppose slope of midparent-offspring regression is .75.
- Hence, h2 = 0.75.
- If a parent is 10 units below the mean, then the average value of its offspring is (.75/2)*10 = 3.75 units below the mean
- Suppose the average value of both parents is 20 units above the mean. Then the average value of their offspring is 0.75*20 = 15 units above the mean.
Heritabilities and Breeding Values
The breeding value of an individual is not something that we can directly observe. We can, however estimate it from an individual's phenotype:
A = h2(z-u)
Thus if Fred's phenotype is 30 and the population mean is 10, Fred is 20 units over the mean. The estimate of Fred's breeding value for this character is
h2*20.
Lecture 48