600A Lecture 28: Quantitative Genetics II. The Resemblance between Relatives
(Current Version: 29 April 2003)
This material is copyrighted and MAY NOT be used for commercial purposes
| You are visitor number |
 |
since 17 April 2003 |
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 27) 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.
Response to Selection
R = h2 S
- R = change in population mean (from one generation to the next)
- S = Mean selected parents - population mean
S is called the Selection differential
Example If height in reproducing parents is 75 inches, while the population average is 65,
- S = 75-65 = 10
- R = h2* 10
- The new mean becomes 65 + h2*10
What happens when selection occurs only on one sex?
R = (h2/2) S
For example, suppose we are selecting for height in corn, and only let tall individuals become pollinated.
Here, there is no selection on the parents producing the pollens, only those receiving it.
Thus, the heritability of height is 0.5 and we only let plants 12 inches over the mean hieght become pollinated, the expected increase in height is (0.5/2)(12) = 3 inches.
Fisher's Fundamental Theorem
Fisher showed that the rate of selection response is proportional to the additive genetic variance in fitness.
Corollary: At equilibrium, additive variance in fitness is zero.
We can see this by recalling that the marginal fitness of an allele (say A), WA equals the average effect of that allele on fitness. Likewise, the variance of average effects is just the additive variance.
If all alleles have the same marginal fitness, there is no change in allele frequencies. Likewise, there is no additive variance as all the average effects (here the marginal finesses) are identical.
Example
Suppose the genotypes AA : Aa : aa have fitnesses 1: 1+s : 1. This is overdominant selection with an equilibrium allele frequnecy of freq(A) = 1/2.
What happens to the heritability of fitness?
Thus, at the equilibrium value the heritability equals zero.
What happens to the genetic variances?
Note that p = 1/2 corresponds to a stable equilibrium point, but that the total genetic variance is largest at this point. It is only the ADDITIVE variance that is zero. This is one reason we worry so much about what fraction of the genetic variance is additive.
Fisher's Fundamental Theorem applies to FITNESS, not the other characters.
Suppose the genotypes AA : Aa : aa have average heights of z = 15: 20: 25, but that selection acts on height, with
W(z) = 1- 0.01 (z -20)2
This is an example of stabilizing selection
Heritability (and additive genetic variance) in height is maximized at p = 1/2, but heritability in FITNESS is zero at this value (as the fitnesses are 0.75: 1 : 0.75).
Fitness surfaces, W(z)
The Fitness surface, W(z) maps character values z into fitnesses. We have seen an example with stabilizing selection above. Fitness surfaces can also show directional selection and disruptive selection.
The local geometry of the fitness surfaces informs us as to what type of selection the character is experiencing.
Fitness surfaces can be complex, showing regions of stabilizing selection, regions of directional selection, and regions of distributive selection.
Selection tends to act in such as way as to increase the fitness of the population, so that characters evolve to increase fitness.
Long- term selection response
Our expression R = h2S predicts the single-generation response to selection. What happens after multiple generations of selection?
The rate of response eventually declines and approaches an apparent selection limit as allele frequency changes cause the additive variance in fitness to approach zero.
Note that the heritability can actually increase before it starts to decline.
Does the selection response really grind to such a halt? Yes for the short-term. Further response occurs as new variation is generated by mutation.