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Lecture 8 Short-Term Selection Response R = h 2 S.

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1 Lecture 8 Short-Term Selection Response R = h 2 S

2 Applications of Artificial Selection Applications in agriculture and forestry Creation of model systems of human diseases and disorders Construction of genetically divergent lines for QTL mapping and gene expression (microarray) analysis Inferences about numbers of loci, effects and frequencies Evolutionary inferences: correlated characters, effects on fitness, long-term response, effect of mutations

3 Response to Selection Selection can change the distribution of phenotypes, and we typically measure this by changes in mean –This is a within-generation change Selection can also change the distribution of breeding values –This is the response to selection, the change in the trait in the next generation (the between- generation change)

4 The Selection Differential and the Response to Selection The selection differential S measures the within-generation change in the mean –S =  * -  The response R is the between-generation change in the mean –R(t) =  (t+1) -  (t)

5 Truncation selection uppermost fraction p chosen Within-generation change Between-generation change

6 The Breeders’ Equation: Translating S into R ( ) Recall the regression of offspring value on midparent value Averaging over the selected midparents, E[ (P f + P m )/2 ] =  *, E[ y o -  ] = h 2 (  -  ) = h 2 S Likewise, averaging over the regression gives Since E[ y o -  ] is the change in the offspring mean, it represents the response to selection, giving: R = h 2 SThe Breeders’ Equation

7 Note that no matter how strong S, if h 2 is small, the response is small S is a measure of selection, R the actual response. One can get lots of selection but no response If offspring are asexual clones of their parents, the breeders’ equation becomes – R = H 2 S If males and females subjected to differing amounts of selection, – S = (S f + S m )/2 –An Example: Selection on seed number in plants -- pollination (males) is random, so that S = S f /2

8 Break for Problem 1

9 Response over multiple generations Strictly speaking, the breeders’ equation only holds for predicting a single generation of response from an unselected base population Practically speaking, the breeders’ equation is usually pretty good for 5-10 generations The validity for an initial h 2 predicting response over several generations depends on: –The reliability of the initial h 2 estimate –Absence of environmental change between generations –The absence of genetic change between the generation in which h 2 was estimated and the generation in which selection is applied

10 50% selected V p = 4, S = % selected V p = 4, S = % selected V p = 1, S = 1.4 The selection differential is a function of both the phenotypic variance and the fraction selected

11 The Selection Intensity, i As the previous example shows, populations with the same selection differential (S) may experience very different amounts of selection The selection intensity i provided a suitable measure for comparisons between populations,

12 Selection Differential Under Truncation Selection Mean of truncated distribution Change in the Variance Height of the unit normal density function at the truncation point T P ( z > T) Phenotypic standard deviations Hence, S =  p T /  T Likewise, i = S/  = p T /  T () ( )

13 Selection Intensity Versions of the Breeders’ Equation We can also write this asSince h 2  P = (  2 A /  2 P )  P =  A (  A /  P )  h  A R = i h  A Since h = correlation between phenotypic and breeding values, h = r PA R = i r PA  A Response = Intensity * Accuracy * spread in Va When we select an individual solely on their phenotype, the accuracy (correlation) between BV and phenotype is h

14 Accuracy of selection More generally, we can express the breeders equation as R = i r uA  A Where we select individuals based on the index u (for example, the mean of n of their sibs). r ua = the accuracy of using the measure u to predict an individual's breeding value = correlation between u and an individual's BV Example: Estimating an individual’s BV with progeny testing. For n progeny, r uA 2 = n/(n+a), where a = (4-h 2 )/h 2 Mathematica program

15 Overlapping Generations R y = i m + i f L m + L f h2ph2p L x = Generation interval for sex x = Average age of parents when progeny are born The yearly rate of response is Trade-offs: Generation interval vs. selection intensity: If younger animals are used (decreasing L), i is also lower, as more of the newborn animals are needed as replacements

16 Generalized Breeder’s Equation R y = i m + i f L m + L f r uA  A Tradeoff between generation length L and accuracy r The longer we wait to replace an individual, the more accurate the selection (i.e., we have time for progeny testing and using the values of its relatives)

17 Finite sample correction for i When a finite number of individuals form the next generation, previous expressions overestimate i Suppose M adults measured, of which N are selected for p = N/M E[i] depends on the expected value of the order statistics Rank the M phenotypes as z 1,M > z 2,M... > z M,M z k,M = kth order statistic in a sample of M ( Standardized order statistics

18 Finite population size results in infinite M expression overestimating the actual selection intensity, although the difference is small unless N is very small.

19 Burrows' approximation (normally-distributed trait) Bulmer’s approximation: uses infinite population result, with p replaced by

20 Selection on Threshold Traits Assume some underlying continuous value z, the liability, maps to a discrete trait. z < T character state zero (i.e. no disease) z > T character state one (i.e. disease) Alternative (but essentially equivalent model) is a probit (or logistic) model, when p(z) = Prob(state one | z)

21 Liability scale Mean liability before selection Mean liability after selection (but before reproduction) Selection differential on liability scale q t * - q t is the selection differential on the phenotypic scale Mean liability in next generation Frequency of character state on in next generation Frequency of trait

22 Steps in Predicting Response to Threshold Selection i) Compute initial mean  0 We can choose a scale where the liability z has variance of one and a threshold T = 0 Hence, z -  0 is a unit normal random variable P(trait) = P(z > 0) = P(z -  > -  ) = P(U > -  ) U is a unit normal Define z [q] = P(U z [1-q] ) = q For example, suppose 5% of the pop shows the trait. P(U > 1.645) = 0.05, hence  = General result:  = - z [1-q] ii) The frequency q t+1 of the trait in the next generation is just q t+1 = P(U > -  t+1 ) = P(U > - [h 2 S +  t ] ) = P(U > - h 2 S - z [1-q] )

23 * iii) Hence, we need to compute S, the selection differential on liability Let p t = fraction of individuals chosen in generation t that display the trait -> This fraction does not display the trait, hence z < 0 This fraction display the trait, hence z > 0 When z is normally distributed, this reduces to Height of the unit normal density function At the point  t Hence, we start at some initial value given h 2 and  0, and iterative to obtain selection response

24 Generation Selection differential S q, Frequency of character Sq Initial frequency of q = Selection only on adults showing the trait

25 15 Minute Break

26 Permanent Versus Transient Response Considering epistasis and shared environmental values, the single-generation response follows from the midparent-offspring regression ( ) Breeder’s Equation Response from epistasis Response from shared environmental effects Permanent component of response Transient component of response --- contributes to short-term response. Decays away to zero over the long-term

27 Response with Epistasis ( ) ( ) The response after one generation of selection from an unselected base population with A x A epistasis is The contribution to response from this single generation after  generations of no selection is c is the average (pairwise) recombination between loci involved in A x A Contribution to response from epistasis decays to zero as linkage disequilibrium decays to zero Response from additive effects (h 2 S) is due to changes in allele frequencies and hence is permanent. Contribution from A x A due to linkage disequilibrium

28 Why unselected base population? If history of previous selection, linkage disequilibrium may be present and the mean can change as the disequilibrium decays More generally, for t generation of selection followed by  generations of no selection (but recombination)

29 Maternal Effects: Falconer’s dilution model z = G + m z dam + e Direct genetic effect on character G = A + D + I. E[A] = (A sire + A dam )/2 Maternal effect passed from dam to offspring is just A fraction m of the dam’s phenotypic value m can be negative --- results in the potential for a reversed response The presence of the maternal effects means that response is not necessarily linear and time lags can occur in response

30 Parent-offspring regression under the dilution model In terms of parental breeding values, Regression of BV on phenotype With no maternal effects, b az = h 2 With maternal effects, a covariance between BV and maternal effect arises, with The resulting slope becomes b Az = h 2 2/(2-m) The response thus becomes ( )

31 Generation Cumulative Response to Selection (in terms of S) Response to a single generation of selection Reversed response in 1st generation largely due to negative maternal correlation masking genetic gain Recovery of genetic response after initial maternal correlation decays h 2 = 0.11, m = (litter size in mice)

32 Generation Cumulative Response (in units of S) m = m = -0.5 m = h 2 = 0.35 Selection occurs for 10 generations and then stops

33 Gene frequency changes under selection Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Fitnesses 1 1+s1+2s Additive fitnesses Let q = freq(A 2 ). The change in q from one generation of selection is:  In finite population, genetic drift can overpower selection. In particular, when drift overpowers the effects of selection

34 Strength of selection on a QTL Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Contribution to Character 0 a2a Have to translate from the effects on a trait under selection to fitnesses on an underlying locus (or QTL) Suppose the contributions to the trait are additive: For a trait under selection (with intensity i) and phenotypic variance  P 2, the induced fitnesses are additive with s = i (a /  P ) Thus, drift overpowers selection on the QTL when

35 More generally Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Contribution to trait 0a(1+k)2a Fitness 11+s(1+h)1+2s  Change in allele frequency: s = i (a /  P ) Selection coefficients for a QTL h = k

36 15 Minute Break

37 Changes in the Variance under Selection The infinitesimal model --- each locus has a very small effect on the trait. Under the infinitesimal, require many generations for significant change in allele frequencies However, can have significant change in genetic variances due to selection creating linkage disequilibrium Under linkage equilibrium, freq(AB gamete) = freq(A)freq(B) With positive linkage disequilibrium, f(AB) > f(A)f(B), so that AB gametes are more frequent With negative linkage disequilibrium, f(AB) < f(A)f(B), so that AB gametes are less frequent

38 * Changes in V A with disequilibrium Under the infinitesimal model, disequilibrium only changes the additive variance. Starting from an unselected base population, a single generation of selection generates a disequilibrium contribution d to the additive variance Additive genetic variance after one generation of selection Additive genetic variance in the unselected base population. Often called the additive genic variance disequilibrium Changes in V A changes the phenotypic variance Changes in V A and V P change the heritability The amount disequilibrium generated by a single generation of selection is Within-generation change in the variance A decrease in the variance generates d < 0 and hence negative disequilibrium An increase in the variance generates d > 0 and hence positive disequilibrium

39 A “Breeders’ Equation” for Changes in Variance d(0) = 0 (starting with an unselected base population) Decay in previous disequilibrium from recombination New disequilibrium generated by selection that is passed onto the next generation Many forms of selection (e.g., truncation) satisfy k > 0. Within-generation reduction in variance. negative disequilibrium, d < 0 k < 0. Within-generation increase in variance. positive disequilibrium, d > 0 d(t+1) - d(t) measures the response in selection on the variance (akin to R measuring the mean) Within-generation change in the variance -- akin to S

40 Break for Problem 2

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