Mapping Phenotypic Plasticity of a Count Trait

Arthur BergDepartment of Statistics, University of Florida

Caption: Marsh plant (Sagittaria sagittifolia) that is(a) partially submerged, (b) completely terrestrial, (c) completely submerged.From Developmental Plasticity and Evolution by David W. Pfennig

Introduction The Model EM algorithm Hypothesis Tests Results

Phenotypic PlasticityDefinitionPhenotypic plasticity is the ability of a genotype to produce differentphenotypes in response to changing environmental conditions (Bradshaw, 1965).

Bradshaw AD (1965). Evolutionary significance of phenotypic plasticity inplants. Adv Genet 13: 115-155.

Caption: Desert locus (Schistocerca gregaria) exhibiting a form of phenotypicplasticity known as phase polyphenism.

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Introduction The Model EM algorithm Hypothesis Tests Results

Visualizing Plasticity with Reaction Norms

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Jumping Ahead—Significant QTL Detected

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Introduction The Model EM algorithm Hypothesis Tests Results

Some Hypotheses for Phenotypic Plasticity

These hypotheses are NOT necessarily mutually exclusive.

Overdominance Hypothesis — Homeostasis

Pleiotropic Hypothesis — Allelic SensitivityEpistatic Hypothesis — Gene RegulationSpecialization Hypothesis — Ecotypic Adaptation

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Introduction The Model EM algorithm Hypothesis Tests Results

Kirst Poplar ExperimentDOE Grant: Genomic Mechanisms of Carbon Allocation and Partitioning in Poplar

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Introduction The Model EM algorithm Hypothesis Tests Results

Discrete Measurement – Sylleptic Branches

Characteristics of the DataDiscrete (count trait)

Three clones per progeny

Two treatments

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Introduction The Model EM algorithm Hypothesis Tests Results

Branch Numbers

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Introduction The Model EM algorithm Hypothesis Tests Results

Literature Review – PlasticityCount Trait Plasticity

Norga et. al., 2003. Quantitative analysis of bristle number in Drosophila mutants identifies genes involved in neural

development. Curr. Biol. 13: 1388-1397.

Marron et. al., 2006. Plasticity of growth and sylleptic branchiness in two poplar families grown at three sites across

Europe. Tree Physiology 26: 935-946.

Plasticity in PopulusWu, R. L., 1998. The detection of plasticity genes in heterogeneous environments. Evolution 52: 967-977.

Wu, R., and R. F. Stettler, 1998. Heredity 81: 299-310.

Plasticity in DrosophilaLeips, J., and T. F. C. Mackay, 2000 Quantitative trait loci for lifespan in Drosophila melanogaster: Interactions with

genetic background and larval density. Genetics 155: 1773-1788.

Plasticity in ArabidopsisKliebenstein et. al., 2002 Genetic architecture of plastic methyl jasmonate responses in Arabidopsis thaliana. Genetics

161: 1685-1696.

Plasticity in C. elegansGutteling et. al., 2007. Mapping phenotypic plasticity and genotype× environment interactions affecting life-history

traits in Caenorhabditis elegans. Heredity 98: 28-37.Arthur Berg Mapping Phenotypic Plasticity of a Count Trait 9/ 18

Introduction The Model EM algorithm Hypothesis Tests Results

Literature Review – Mapping QTL With Count DataMaximum Likelihood

Rebaï, A., 1997. Comparison of methods for regression intervalmapping in QTL analysis with non-normal traits. Genetics69:69-74.

Least-Squares RegressionShepel, L. A., H. Lan, J. D. Haag, G. M. Brasic, M. E. Gheen et

al., 1998. Genetic identification of multiple loci that control breastcancer susceptibility in the rat. Genetics 149: 289-299.

Bayesian FrameworkSen, S., and G. A. Churchill, 2001. A statistical framework for

quantitative trait mapping. Genetics 159: 371-387.Poisson Regression

Yuehua Cui, Dong-Yun Kim, and Jun Zhu, 2006. On theGeneralized Poisson Regression Mixture Model for MappingQuantitative Trait Loci With Count Data. Genetics 174: 2159-72.

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Introduction The Model EM algorithm Hypothesis Tests Results

Multivariate Poisson Mixture Model

Subscriptsr = 1, . . . ,R — replicates

k = 1, 2 — treatments

i = 1, . . . , n — progeny

j = 1, 2 — QTL genotype

Xik = (X1ik, ...,XRik) ∼ P(Xik|Θk) = ω1|iP1(Xik|Θ1|k) + ω0|iP0(Xik|Θ0|k),

Pj(Xik|Θj|k) = exp

(−

R∑r=1

λj|rk

)R∏

r=1

λXrikj|rk

Xrik!

sik∑r=0

R∏l=1

(Xlik

i

)r!

(λj|0k∏R

r=1 λj|rk

)r

sik = min(X1ik, ...,XRik)λj|0k + λj|rk is the mean count trait for QTL genotype j in replicate r

λj|0k is the QTL genotype-specific covariance between all pairs of the counttrait in different replicates

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Introduction The Model EM algorithm Hypothesis Tests Results

Assumptions

trait values from different treatments are independent

L(ω,Θk|Xik,M) =n∏

i=1

[ω1|iP1(Xi1|Θ1|1) + ω0|iP0(Xi1|Θ0|1)

]×

n∏i=1

[ω1|iP1(Xi2|Θ1|2) + ω0|iP0(Xi2|Θ0|2)

] (1)

genotypic means of the count trait are equal over replicates under eachtreatment

λj|0k + λj|1k = ... = λj|0k + λj|Rk = λj|0k + λj|k

or equivalently,λj|1k = ... = λj|Rk = λj|k

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A Two-Stage Hierarchical EM Algorithm – E step

E step:

s(t)1|ik =

P1(Xik − 1|Θ(t)1|k)

P1(Xik|Θ(t)1|k)

,

s(t)0|ik =

P0(Xik − 1|Θ(t)0|k)

P0(Xik|Θ(t)0|k)

,

Ω(t)1|ik =

ω1|iP1(Xik|Θ(t)1|k)

ω1|iP1(Xik|Θ(t)1|k) + ω0|iP0(Xik|Θ(t)

0|k),

Ω(t)0|ik =

ω0|iP0(Xik|Θ(t)0|k)

ω1|iP1(Xik|Θ(t)1|k) + ω0|iP0(Xik|Θ(t)

0|k).

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A Two-Stage Hierarchical EM Algorithm – M step

M step:

λ(t+1)1|0k = λ

(t)1|0k

∑ni=1 Ω(t)

1|iks(t)1|ik∑n

i=1 Ω(t)1|ik

,

λ(t+1)0|0k = λ

(t)0|0k

∑ni=1 Ω(t)

0|iks(t)0|ik∑n

i=1 Ω(t)0|ik

,

λ(t+1)1|k =

∑ni=1 Ω(t)

1|ikXik∑ni=1 Ω(t)

1|ik

− λ(t)1|0k,

λ(t+1)0|k =

∑ni=1 Ω(t)

0|ikXik∑ni=1 Ω(t)

0|ik

− λ(t)0|0k,

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Introduction The Model EM algorithm Hypothesis Tests Results

Hypothesis TestsTesting QTL ExistenceH0 : Θ1|1 = Θ0|1 = Θ1 and Θ1|2 = Θ0|2 = Θ2 vs. H1 : Not H0

Hypothesis Test (Testing Pleiotropic Effect – Treatment I)H0 : Θ1|1 = Θ0|1 = Θ1 vs. H1 : Not H0

Hypothesis Test (Testing Pleiotropic Effect – Treatment II)H0 : Θ1|2 = Θ0|2 = Θ2 vs. H1 : Not H0

Hypothesis Test (Testing Genotype by Environment Interaction)H0 : Θ1|1 −Θ0|1 = Θ1|2 −Θ0|2 vs. H1 : Not H0

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Introduction The Model EM algorithm Hypothesis Tests Results

Significant QTL Detected

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Introduction The Model EM algorithm Hypothesis Tests Results

Hypothesis Tests on the Significant QTL

Low Fertilization:p = .05High fertilization:p = .02Interaction:p = .03

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Thank you!