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**Lecture 8 Short-Term Selection Response**

R = h2 S

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**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

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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)

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**The Selection Differential and the Response to Selection**

The selection differential S measures the within-generation change in the mean S = m* - m The response R is the between-generation change in the mean R(t) = m(t+1) - m(t)

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

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**The Breeders’ Equation: Translating S into R**

Recall the regression of offspring value on midparent value ( ) Averaging over the selected midparents, E[ (Pf + Pm)/2 ] = m*, E[ yo - m ] = h2 ( m* - m ) = h2 S Likewise, averaging over the regression gives Since E[ yo - m ] is the change in the offspring mean, it represents the response to selection, giving: R = h2 S The Breeders’ Equation

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**If males and females subjected to differing amounts of selection,**

Note that no matter how strong S, if h2 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 = H2 S If males and females subjected to differing amounts of selection, S = (Sf + Sm)/2 An Example: Selection on seed number in plants -- pollination (males) is random, so that S = Sf/2

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Break for Problem 1

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**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 h2 predicting response over several generations depends on: The reliability of the initial h2 estimate Absence of environmental change between generations The absence of genetic change between the generation in which h2 was estimated and the generation in which selection is applied

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**The selection differential is a function of both**

the phenotypic variance and the fraction selected 20% selected Vp = 1, S = 1.4 50% selected Vp = 4, S = 1.6 20% selected Vp = 4, S = 2.8

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**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,

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**Selection Differential Under Truncation Selection**

Mean of truncated distribution ( ) Phenotypic standard deviations Height of the unit normal density function at the truncation point T P ( z > T) Hence, S = s pT/pT Likewise, i = S/ s = pT/pT ( ) Change in the Variance

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**Selection Intensity Versions of the Breeders’ Equation**

Since h2sP = (s2A/s2P) sP = sA(sA/sP) = h sA We can also write this as R = i h sA Since h = correlation between phenotypic and breeding values, h = rPA R = i rPAsA Response = Intensity * Accuracy * spread in Va When we select an individual solely on their phenotype, the accuracy (correlation) between BV and phenotype is h

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**Accuracy of selection R = i ruAsA**

More generally, we can express the breeders equation as R = i ruAsA Where we select individuals based on the index u (for example, the mean of n of their sibs). rua = 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, ruA2 = n/(n+a), where a = (4-h2)/h2 Mathematica program

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**Overlapping Generations**

Lx = Generation interval for sex x = Average age of parents when progeny are born The yearly rate of response is Ry = im + if Lm + Lf h2sp 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

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**Generalized Breeder’s Equation**

Ry = im + if Lm + Lf ruAsA 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)

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**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 z1,M > z2,M . .. > zM,M zk,M = kth order statistic in a sample of M ( Standardized order statistics

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**p = N/M, the proportion selected Expected Selection intensity**

1 .1 .01 p = N/M, the proportion selected Expected Selection intensity M = 10 M = 20 M = 50 M = 100 M = infinity Finite population size results in infinite M expression overestimating the actual selection intensity, although the difference is small unless N is very small.

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**Burrows' approximation (normally-distributed trait)**

Bulmer’s approximation: uses infinite population result, with p replaced by

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**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)

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

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**Steps in Predicting Response to Threshold Selection**

i) Compute initial mean m0 Hence, z - m0 is a unit normal random variable U is a unit normal P(trait) = P(z > 0) = P(z - m > -m) = P(U > -m) We can choose a scale where the liability z has variance of one and a threshold T = 0 Define z[q] = P(U < z[q] ) = q. P(U > z[1-q] ) = q For example, suppose 5% of the pop shows the trait. P(U > 1.645) = 0.05, hence m = General result: m = - z[1-q] ii) The frequency qt+1 of the trait in the next generation is just qt+1 = P(U > - mt+1 ) = P(U > - [h2S + mt ] ) = P(U > - h2S - z[1-q] )

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**iii) Hence, we need to compute S, the selection**

differential on liability Let pt = 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 mt * - Hence, we start at some initial value given h2 and m0, and iterative to obtain selection response

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**Initial frequency of q = 0.05. Selection only on adults **

showing the trait 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 10 20 30 40 50 60 70 80 90 100 S q q, Frequency of character Selection differential S 25 20 15 10 5 Generation

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15 Minute Break

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**Permanent Versus Transient Response**

Considering epistasis and shared environmental values, the single-generation response follows from the midparent-offspring regression ( ) 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 Breeder’s Equation Response from epistasis

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**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 t generations of no selection is ( ) Response from additive effects (h2 S) is due to changes in allele frequencies and hence is permanent. Contribution from A x A due to linkage disequilibrium 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

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**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 t generations of no selection (but recombination)

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**Maternal Effects: Falconer’s dilution model z = G + m zdam + e**

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

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**) ( Parent-offspring regression under the dilution model**

In terms of parental breeding values, Regression of BV on phenotype With no maternal effects, baz = h2 With maternal effects, a covariance between BV and maternal effect arises, with The resulting slope becomes bAz = h2 2/(2-m) The response thus becomes ( )

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**Response to a single generation of selection**

h2 = 0.11, m = (litter size in mice) Recovery of genetic response after initial maternal correlation decays -0.15 -0.10 -0.05 0.00 0.05 0.10 Cumulative Response to Selection Reversed response in 1st generation largely due to negative maternal correlation masking genetic gain (in terms of S) 10 9 8 7 6 5 4 3 2 1 Generation

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**Selection occurs for 10 generations and then stops**

-1.0 -0.5 0.0 0.5 1.0 1.5 m = -0.25 m = -0.5 m = -0.75 Cumulative Response (in units of S) h2 = 0.35 20 15 10 5 Generation

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**Gene frequency changes under selection**

Genotype A1A1 A1A2 A2A2 Fitnesses 1 1+s 1+2s Additive fitnesses Let q = freq(A2). The change in q from one generation of selection is: D In finite population, genetic drift can overpower selection. In particular, when drift overpowers the effects of selection

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**Strength of selection on a QTL**

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: Genotype A1A1 A1A2 A2A2 Contribution to Character a 2a For a trait under selection (with intensity i) and phenotypic variance sP2, the induced fitnesses are additive with s = i (a /sP ) Thus, drift overpowers selection on the QTL when

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**D More generally Change in allele frequency:**

Genotype A1A1 A1A2 A2A2 Contribution to trait a(1+k) 2a Fitness 1 1+s(1+h) 1+2s D Change in allele frequency: s = i (a /sP ) Selection coefficients for a QTL h = k

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15 Minute Break

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**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 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 Under linkage equilibrium, freq(AB gamete) = freq(A)freq(B)

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**Changes in VA with disequilibrium**

Starting from an unselected base population, a single generation of selection generates a disequilibrium contribution d to the additive variance Under the infinitesimal model, disequilibrium only changes 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 VA changes the phenotypic variance The amount disequilibrium generated by a single generation of selection is Changes in VA and VP change the heritability Within-generation change in the variance An increase in the variance generates d > 0 and hence positive disequilibrium A decrease in the variance generates d < 0 and hence negative disequilibrium *

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**A “Breeders’ Equation” for Changes in Variance**

d(0) = 0 (starting with an unselected base population) Within-generation change in the variance -- akin to S New disequilibrium generated by selection that is passed onto the next generation Decay in previous disequilibrium from recombination d(t+1) - d(t) measures the response in selection on the variance (akin to R measuring the mean) 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

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Break for Problem 2

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