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Evolution of resistance to white pine blisterrust in high-elevation pines

Mike Antolin, Stuart Field, Jen Klutsch, Anna Schoettle and Simon Tavener

Colorado State University

Rocky Mountain Research Station, USDA Forest Service

Thanks: NSF award 0734267, USDA Forest Service

Evolution of resistance – p. 1/74

The basic ecology

High-elevation White Pines

• Habitat: dry, exposed, high altitude sites (Rocky Mtns.)

• Occupies habitat other species cannot

• Long-lived species

• Keystone species

Threats to HEWPs

White pine blister rustClimate change

Mountain Pine Beetle

White Pine Blister Rust

• Cronartium ribicola

• Non-native fungal pathogen

• Lethal in HEWP

• Spreading despite extensive control efforts

Effects of WPBR

WPBR Life Cycleaeciospores

basidiospores

Two Obligate Hosts

• Ribes spp.• Spring• Sub-lethal

• HEWP• Fall• Lethal

B = Basic model

B: Basic n-stage model

Consider a stage-structured population model in the form of a nonlinear map

x{n}(t+ 1,p) = h{n}(x{n}(t,p),p)

x{n}(0,p) = x{n},0

h{n} : Rn × Rm 7→ R

Notation

• The subscript {n} is used to denote the size of the system.

• x{n}(t,p) is the solution vector containing xi, the density of individuals

(individuals per hectare) in stage i = 1, . . . , n at times t = 0, 1, . . . , T

(years).

• p is the vector of parameters pj , j = 1, . . . ,M .

Nonlinear o.d.e. models are analogous.Evolution of resistance – p. 4/74

B: Order of operations

Assumption: Fecundity occurs after survivorship & transition.

Consequently,

x{n}(t+ 1) = g{n}(x{n}(t)) + f{n}

(g{n}(x{n}(t))

where f denotes fecundity and g denotes survivorship and transition.

Note: When considering a genetic model, a splitting between survivorship &

transition and fecundity arises naturally since survivorship & transition occurswithin genotypes while fecundity occurs across genotypes.

Note: For simplicity, we drop the explicit dependence of solution x{n}(t) on

parameters p.

Notation

We will use

• x to denote the outcome of survivorship & transition,

• y as the vector on which fecundity operates,

• x or y to denote intermediate quantities.

Assumption: Disease status and genotype affect the arguments (inputs) of thefunctions modeling survivorship & transition and fecundity.

D = Disease

D: Incorporating disease

Assumption: There are only susceptible and infected populations.

x{2n} =

g{2n}(x{2n}) =

g{n}(xS{n})

g{n}(xI{n})

White bark pines do not recover from blister rust infection

D: Survivorship & transition

Assumption: The effect of disease can be described in terms of a linear

weighting applied to the argument of the function modeling survivorship &transition.

Let the cost of infection to survivorship & transition be

Cν{2n} =

I{n} 0

0 Cν{n}

, Cν{n} = diag(ci), i = 1, . . . , n ,

and define

x{2n} = g{2n}

ν{2n} ∗ x{2n}

g{n}(xS{n})

g{n}(Cν{n} ∗ xI

D: Fecundity

Fecundity is modeling through the action of the non-linear vector valued

function

f{2n} : R2n 7→ R2n,

f{2n} 6=

f{n}(xS{n})

f{n}(xI{n}).

Think vertical vs horizontal transmission.

Both susceptible and infected white bark pine produce susceptible seeds.

(Furthermore, seeds cannot become infected.)

D: Fecundity

Assumption: The effect of disease on fecundity can be described as a

weighting applied to the argument of the nonlinear function used to modelfecundity.

Let the cost of fecundity be

{2n} =

and define fecundity in the SI model to be

{2n} ∗ x{2n}

D: Infectivity

Assumption: Infection occurs after survivorship, transition and fecundity and

occurs at a constant rate.

B{2n} =

I{n} −B{n} 0

B{n} I{n}

B{n} = diag(bi), i = 1, . . . , n

and I{n} is the n× n identity matrix.

D: SI disease model

Finally, the full SI disease model is the nonlinear map

x{2n}(t+ 1) = B{2n} ∗(x{2n} + f{2n}(y{2n})

x{2n} = g{2n}

ν{2n} ∗ x{2n}

y{2n} = Cϕ

{2n} ∗ x{2n}.

D: SI disease model

The cost of disease on survival, transition, the cost of disease on fecundity and

infectivity could all be nonlinear.

The SI disease model would then become

x{2n}(t+ 1) = B{2n}

(x{2n} + f{2n}(y{2n})

x{2n} = Cν{2n}

(x{2n}

y{2n} = Cϕ

(x{2n}

Conceptually this is no more complicated.

P = Pine model

P: Variables

The basic pine model has six stages or classes,

x{6} =

Units: Individuals / hectare

P: The nonlinear map

Consistent with the general framework, we define the nonlinear map

x{6}(t+ 1) = g{6}(x{6}(t)) + f{6}(g{6}(x{6}(t)))

where f and g represent fecundity and survivorship & transition respectively.

Units: Time is measured in years.

Model structure and coefficient values is the result of extensive field studies.

P: Nonlinear seedling recruitment

Assumption: Seedling recruitment occurs before survivorship & transition

The modeling of the germination process involves a number of auxiliary

quantities, including properties of the surrounding forest, and the foraging andseed storing behaviors of the Clark’s nutcracker.

• Leaf Area Index (LAI)

• Seed caching

• Pfind

• . . .

P: Seedling recruitment. Leaf Area Index

Leaf Area Index (LAI) is used as a proxy for tree density and is defined as a

function of tree size as measured by the diameter at breast height.

larea1= larea2

= 0, larea3= α1, lareai

= α2 (di)α3 , i = 4, 5, 6,

α1 = 0.456, α2 = 0.117, α3 = 1.925 ,

and the diameter at breast height (dbh) measurements for the six classes are

d1 = 0, d2 = 0, d3 = 0, d4 = 2.05, d5 = 12.5, d6 = 37.

Secondary seedlings have d3 = 0 but do possess leaf area, thus ~larea3is

calculated separately. Let

LAI(x6) =(larea, x6)

10000.

P: Seedling recruitment. Clark’s nutcracker

We define

SpB(x{6}) =x1

nBirds,

rcache(x{6}) =0.73

1 + exp((31000− SpB(x{6}))/3000

) + 0.27 ,

rALs(x{6}) =1

1 + exp(2 (LAI(x{6})− 3)),

where x1 is the number of seeds, and finally,

r2(x{6}) =

[(1− Pfind) (1− Pcons)

]rcache(x{6}) rALs(x{6}) .

The number of new seedlings is then given by

γ{6}(x{6}) = r2(x{6}) ∗ x1 ∗ e2 ,

where e2 ∈ R6 is the unit vector with a single nonzero entry in the second

position.Evolution of resistance – p. 20/74

P: Seedling recruitment

We define the intermediate vector

x{6} = x{6} + γ{6}(x{6}),

where γ{6}(x{6}) is the result of the nonlinear seedling recruitment process.

P: Linear survivorship & transition

S{6} =

1 0 0 0 0 0

0 s2 0 0 0 0

0 t2 s3 0 0 0

0 0 t3 s4 0 0

0 0 0 t4 s5 0

0 0 0 0 t5 s6

The coefficients along the diagonal of S{6} are

s2 = 0.636, s3 = 0.8391, s4 = 0.9310, s5 = 0.9653, s6 = 0.995 .

Coefficients along the sub-diagonal of S{6} are

t2 = 0.212, t3 = 0.0559, t4 = 0.0490, t5 = 0.0197 .

P: Survivorship & transition

We define the outcome of both nonlinear and linear survivorship and transition

x{6} = g{6}(x{6}) = S{6} ∗ x{6}.

P: Fecundity

We define the nonlinear function

f{6} : R6 × ρ 7→ R6

where ρ is a measure of the effect of class on fecundity.

The modeling of the fecundity process involves a number of auxiliaryquantities.

• Maximum number of cones per tree

• Number of seeds per cone

• . . .

P: Fecundity

Seed production is determined by the number of seeds per cone, which is

determined as

Ctree(x{6}) =

1 + exp(5 (LAI(x{6})− 2.25))+ 0.5

r1(x{6}) = Scone ∗ Ctree(x{6}).

The number of seeds produced is then given by

f{6} = r1(x{6}) ∗ (ρx5 + x6) ∗ e1

where e1 ∈ R6 is a unit vector with a single nonzero entry in the first position.

P: The nonlinear map

Finally

x{6}(t+ 1) = x{6} + f{6}(x{6}).

PD = Pine + disease

PD: Variables

We double the number of classes and define

x{12} =

susceptible SD1

susceptible SD2

susceptible SA

susceptible YA

susceptible MA

infected SD1

infected SD2

infected SA

infected YA

infected MA

1) Seedling recruitment : r2

x {n}( )

2) Survival& transition : g {n}

3) Seed production : f {n}

4) Infection : B{n}

Spring

iSPiYA

Arrows:Survival & transitionInfectionFecundity

Nodes:SusceptibleInfected

PD: Seedling recruitment

Assumption: Seedling recruitment occurs before survivorship & transition.

We define

x{12} =

γD{6}(x{12})

where the nonlinear function γD{6} is the modification of γ{6}.

Both susceptible and diseased individuals provide shading.

PD: Survivorship & transition

S{12} = diag(S{6},S{6}).

The cost of infection on survival & transition is given by

Cν{12} = diag(I{6},C

ν{6})

Cν{6} = diag(0, c2, . . . , c6),

with the cost of infection related to tree size, specifically

c2 = 0.01, c3 = 0.13, ci = 1− exp(−δ di), i = 4, 5, 6.

PD: Survivorship & transition

We define

x{12} = g{12}(x{12})

= S{12} ∗(C

ν{12} ∗ x{12}

S{6} 0

0 S{6}

I{6} 0

0 Cν{6}

S{6} 0

0 S{6}Cν{6}

PD: Fecundity

Assumptions:

• Infection is not transmitted vertically. Susceptible and infected adults

produce susceptible seeds.

• Fecundity is modeled via cone production only because we assumefertilization occurs via a “pollen cloud", with each individual contributing

equally to the cloud. Thus pollen is not limiting and differences in fecunditycan be seen as a result of differences in available resources allocated to

cone production.

• Pollen production is unaffected by infection because the pollen producingbranches primarily compose the lower crown, while the seed cone

producing branches dominate the upper crown. Infectiondisproportionately affects the upper crown (top-kill), thus mitigating the

effect of infection on pollen relative to cone production.

PD: Fecundity

f{12} : R12 × (ρ, Cϕ)⊤ 7→ R12

• ρ is a measure of the effect of class on fecundity,

• Cϕ is a measure of the effect of disease on fecundity.

PD: Fecundity

Let y{12} = x{12} and

yS = ρy5 + y6, yI = ρy11 + y12,

z = yS + CϕyI ,

so that

f{12}(x{12}; ρ,Cϕ) = r1(x{12}) ∗ z ∗ e1,

where the function r1(x{12}) is an obvious extension of r1(x{6}). Thinkshading.

PD: Infectivity

Assumption: Infectivity occurs at a constant rate which may vary according to

age class.

We define

B{12} =

I{6} −B{6} 0

B{6} I{6}

B{6} = diag(0, b2, b3, b4, b5, b6)

b2 = b3 = b4 = b5 = b6 = 0.044.

PD: The nonlinear map

Assumption: Disease transmission occurs after fecundity.

Finally

x{12}(t+ 1) = B{12} ∗(x{12} + f{12}(x{12}; ρ, C

x{12} = g{12}(x{12}).

Sensitivity analysis

Sensitivity wrt parameters

Using index notation,

xi(t+ 1,p) = hi(x(t,p), p),

xi(0) = zi

i = 1, . . . , n

Differentiating with respect to parameters, pk, gives

∂xi(t+ 1)

∂pk=

∗∂xm(t)

∂pk+

∂xi(0)

∂pk= 0

i = 1, . . . , n,

k = 1, . . . ,K .

Elasticities are relative sensitivities,

Ei,k =pkxi

∂pk.

Sensitivity wrt initial conditions

To determine stability with respect to the initial conditions, we differentiate with

respect to the initial conditions to give

∂xi(t+ 1)

∂zk=

∗∂xm(t)

∂zk+

∂xi(0)

∂zj= δij

i = 1, . . . , n,

k = 1, . . . ,K ,

To solve this for the population and its stability with respect to parameters andinitial conditions we evolve all equations simultaneously.

Perform calculations using the MATLAB / MAPLE package SENSAI⋆.⋆Methods in Ecology and Evolution, 2, pp 560–575, (2011).

Results

Model Results IRegeneration from 1000 SD1

• Stages 2-6

• Eqm = 626

• No disease (β = 0)

0 200 400 600 800

SD1SD2SAYAMA

Equilibrium Structure

Model Results II

0 200 400 600 800

SD1SD2SAYAMA

Equilibrium Structure: susceptible

0 200 400 600 800

SD1SD2SAYAMA

Equilibrium Structure: infected

• Introduce rust into Eqm population

• Infection scenario (β = 0.044)

The Beta Effect

0 50 100 150 200

! = 0.2

0 50 100 150 200

SD1SD2SAYAMA

0 50 100 150 200

! = 0.016

0 50 100 150 200

! = 0.044

0 0.016

0.0440.2

Beta vs. Delta

Total Trees - 100 yrs 500

Mature Adults - 100 yrs 0

Mature AdultsTotal Trees

100 years

Disease free sensitivities and elasticities

Mature adults and total population.Regenerating population after 100 years and equilibrium population.

Elasticities during transience

Elasticities for β = 0.044 after 100 years.

Mature adult population and total population.

Infectivity vs Prevalence of disease

Stage-specific and overall prevalence of blister rust for low (β = 0.016),

medium (β = 0.044) and high (β = 0.20) probability of infection.

Conclusions I

• Sustainability of HEWP stands infected with WPBR depends on twodominant effects:

(a) Infection probability

(b) Regeneration mediated by competition (e.g. LAI)

Parameters controlling these effects disproportionately remove smallerstages via infection induced mortality and reduced seedling establishment.

• Stand structure

Diseased equilibrium stands are less dominated by mature adults thandisease-free equilibrium stands.

• Disease prevalence

(a) Low prevalence in SD1 is due to high cost of infection, high natural

mortality and low residence time, not low susceptibility. This mayaccount for low seedling rust prevalence found in field surveys.

(b) High prevalence in mature adults due to low infection cost and low

natural mortality. Consistent with field observations.Evolution of resistance – p. 44/74

Conclusions II

• Transient vs equilibrium sensitivity

(a) 100 years after onset of infection, MA are sensitive to mortality and

infection probabilities of all stages

(a) At equilibrium, MA are sensitive to mortality and infection probabilitiesof the MA stage only

What insights can modeling provide?

Simulation

• Reinterpret existing observations

• Understand change in the structure of the forest

• Understand time scale for loss of forest ( » observation time)

• Understand time scale for control strategies ( » observation time)

• Predict behaviors of planted stands vs mature forests

Transient sensitivities

• Identify important parameters to estimate

• Identify important parameters to seek to control

• Equilibrium sensitivities are somewhat moot

Control strategies

Control efforts

1. Eradicate ribes†

2. Prune infected branches†

3. Select and breed rare resistant individuals

What insights might a genetic model provide?

• Understand time scale for spread of a resistant allele

• Estimate the effect of potential control strategies on pine stands

† = job creation schemeEvolution of resistance – p. 47/74

SLG = Single locus genetics

Single locus genetics

Assumption: Consider two distinct alleles at a single locus only.

x{3n} =

xAA{n}

xaA{n}

xaa{n}

Survivorship & transition occur independently within each genotype, hence

g{3n}(x{3n}) =

g{n}(xAA{n})

g{n}(xaA{n})

g{n}(xaa{n})

SLG: Fecundity

Random mating is applied through the action of the nonlinear vector-valued

function

f{3n} : R3n 7→ R3n.

SLG: The nonlinear map

Assumption: Genotype affects survivorship & transition through fitness

function V{3n} : R3n 7→ R3n and affects fecundity through fitness function

W{3n} : R3n 7→ R3n.

Putting it all together,

x{3n}(t+ 1) = x{3n} + f{3n}(y{3n})

x{3n} = g{3n}

(V{3n}(x{3n})

y{3n} = W{3n}(x{3n}).

DSLG = Disease and single locus genetics

Disease and single-locus genetics

Combining our earlier ideas we define

x{6n} =

xAA{2n}

xaA{2n}

xaa{2n}

DSLG: Survivorship & transition

We extend g{2n} to create g{6n} : R6n 7→ R6n as

g{6n} =

Assumption: Genotype can influence the cost of disease on survivorship &transition.

Cν{6n} = diag

{2n} ,Cν,aA

{2n} ,Cν,aa

= diag(I{n},C

{n} , I{n},Cν,aA

{n} , I{n},Cν,aa

DSLG: Fecundity

We define

f{6n} : R6n 7→ R6n

to model random mating.

Assumption: Genotype can influence the effect of disease on fecundity.

{6n} = diag(C

{2n} ,Cϕ,aA

{2n} ,Cϕ,aa

= diag(I{n},C

{n} , I{n},Cϕ,aA

{n} , I{n},Cϕ,aa

DSLG: Infectivity

Assumption: Infectivity is independent of genotype.

B{6n} = diag(B{2n},B{2n},B{2n}

DSLG: The nonlinear map

x{6n}(t+ 1) = B{6n} ∗(x{6n} + f{6n}(y{6n})

x{6n} = g{6n}(Cν{6n} ∗ x{6n})

y{6n} = Cϕ

{6n} ∗ x{6n}.

PDSLG = Pine + disease + single locusgenetics

PDSLG: Variables

Define x{36} as

x{36} =

xAA{12}

xaA{12}

xaa{12}

Recall that survivorship and transition occurs within each genotype.

PDSLG: Seedling recruitment

Assumption: Nonlinear seedling recruitment is independent of infection status

and genotype.

We define

x{12} = x{12} +

γDSLG{6} (x{36})

x{36} =

xAA{12}

xaA{12}

xaa{12}

Think shading.

PDSLG: Survivorship costs

S{36} = diag(S{6}, S{6}, S{6}, S{6}, S{6}, S{6}),

Cν36 =

... 0 0

0 I6... 0 0

... 0 0

. . . . . . . . . . . . . . . . . . . . . . . .

... I6 0

... 0 0

... 0 I6 − h ∗Cν6

... 0 0

. . . . . . . . . . . . . . . . . . . . . . . .

... 0 0

... I 0

... 0 0

... 0 I6 −Cν6

PDSLG: Survivorship & transition

x{36} = g{36}

(x{36}

= S{36} ∗(C

ν{36} ∗ x{36}

PDSLG: Fecundity

f{36} : R36 × (ρ,Cϕ, h)⊤ 7→ R36

• ρ is a measure of the effect of class on fecundity

• Cϕ is a measure of the effect of disease on fecundity,

• h is the usual partial dominance parameter.

PDSLG: Pollen production

Assumptions

• Males do not have an infection status.

• The number of males linearly dependent on the number of females.

Let y = x{36} and

yAAmale = (y5 + y6) + (y11 + y12),

yaAmale = (y17 + y18) + (y23 + y24),

yaamale = (y29 + y30) + (y35 + y36),

male = yAAmale + yaA

male + yaamale,

so that

ppollen =yAAmale + 0.5 yaA

maleand qpollen =

0.5 yaAmale + yaa

PDSLG: Females

Assumption: Stage has an effect on viability of females that is independent of

infection status and genotype.

Define

yAAS = ρy5 + y6, yAA

I = ρy11 + y12,

yaAS = ρy17 + y18, yaA

I = ρy23 + y24,

ySaa = ρy29 + y30, yaa

I = ρy35 + y36,

female = yAAS + yAA

I + yaAS + yaA

I + yaaS + yaa

PDSLG: Egg viability

Assumption: Genotype affects egg viability of infected individuals.

Consider the effect of infection status and genotype on egg viability

zAA = yAAS + yAA

zaA = yaAS + (1− h ∗ (1− Cϕ) ∗ yaA

zaa = yaaS + Cϕ ∗ yaa

PDSLG: Random mating

Random mating gives

f1(y36) = ppollen ∗ (zAA + 0.5 zaA),

f13(y36) = qpollen ∗ (zAA + 0.5 zaA) + ppollen ∗ (0.5 zaA + zaa),

f25(y36) = qpollen ∗ (0.5 zaA + zaa),

f{36} = f1(y36) ∗ e1 + f13(y36) ∗ e13 + f25(y36) ∗ e25.

f{36} = r1(y36) ∗ f36

where the function r1(x{36}) is an obvious extension of r1(x{6}).

PDSLG: Infectivity

Assumption: Infectivity is independent of genotype.

B{36} =

B{12} 0 0

0 B{12} 0

0 0 B{12}

PDSLG: The nonlinear map

x{36}(t+ 1) = B{36} ∗(x{36} + f{36}

(x{36}; ρ, C

ϕ, h))

Results

Preliminary studies

0 500 1000 15000

1QoI vs Time

LAIb=0, rsite=0.2LAIb=0, rsite=0.4LAIb=0, rsite=1LAIb=2, rsite=0.2LAIb=2, rsite=0.4LAIb=2, rsite=1

0 500 1000 15000

0.5QoI vs Time

LAIb=0, rsite=0.2LAIb=0, rsite=0.4LAIb=0, rsite=1LAIb=2, rsite=0.2LAIb=2, rsite=0.4LAIb=2, rsite=1

Fixation of advantageous allele with and without background pollen

Preliminary studies

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

0 200 400 600 800 1000

Time Step

p = 0.005, LAI.b = 2, r.site=0.4, Beta = 0.044

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0 50 100 150 200

Time Step

ees Quantity of Interest

Parameter

Sensitivity of # Adult Trees, TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

Parameter

Elasticity of # Adult Trees, TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

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●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

0 200 400 600 800 1000

Time Step

p = 0.005, LAI.b = 2, r.site=0.4, Beta = 0.044

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0 50 100 150 200

Time Step

) Quantity of Interest

ParameterS

tivity

Sensitivity of p = f(R), TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

Parameter

Elasticity of p = f(R), TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

Number of adults Frequency of advantageous alleleModerate competition and low quality site.

Preliminary studies

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0 200 400 600 800 1000

Time Step

p = 0.005, LAI.b = 0, r.site=1, Beta = 0.044

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●●●●●

0 50 100 150 200

Time Step

ees Quantity of Interest

Parameter

Sensitivity of # Adult Trees, TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

Parameter

Elasticity of # Adult Trees, TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

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0 200 400 600 800 1000

Time Step

p = 0.005, LAI.b = 0, r.site=1, Beta = 0.044

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

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0 50 100 150 200

Time Step

) Quantity of Interest

ParameterS

tivity

Sensitivity of p = f(R), TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

Parameter

2Elasticity of p = f(R), TimeStep = 200

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33

Number of adults Frequency of advantageous alleleLow competition, high quality site.

Future directions

• Comprehensive study of single locus genetics including control strategies

• Multilocus genetics

1. Linkage disequilibrium for small numbers of loci and alleles

2. Recombination vs selection for small numbers of loci

3. Parametrization ???

Future directions

• Comprehensive study of single locus genetics including control strategies

• Multilocus genetics

1. Linkage disequilibrium for small numbers of loci and alleles

2. Recombination vs selection for small numbers of loci

3. Parametrization ???

There is something fascinating about science. One gets such wholesalereturns of conjecture out of such a trifling investment of fact. ... Mark Twain

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