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Genetic Variability in the Mendelian Diploid Model

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Genetic Variability in the Mendelian Diploid Model

Loren Coquille (Institut Fourier, Grenoble)

joint work in progress with A. Bovier and R. Neukirch (University of Bonn)

Journ´ee de la chaire MMB, 13 avril 2016

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Outline

1 Introduction

2 Clonal reproduction model Trait substitution sequence

3 Mendelian diploid model Genetic variability

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Outline

1 Introduction

2 Clonal reproduction model

3 Mendelian diploid model

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Clonal Reproduction versus Sexual Reproduction

Clonal Reproduction

individual reproduces out of itself

offspring = copy of the parent

offspring genom:

I 100% of genes of the parent

vs. Sexual Reproduction

individual needs a partner for reproduction

offspring genom:

I 50% genes of mother

I 50% genes of father

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Clonal Reproduction versus Sexual Reproduction

Clonal Reproduction

+ fast growing populations + no time loss for partner

selection

- genes stay constant over generations

- hard adaptation to changing environment

Sexual Reproduction

- slowly growing population - time cost for partner selection + genetic variability

+ possible adaptation to changing environment + correction of genetic defects + elimination of disadvantageous

mutations

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Outline

1 Introduction

2 Clonal reproduction model Trait substitution sequence

3 Mendelian diploid model

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Clonal reproduction model

Bolker, Pacala, Dieckmann, Law, Champagnat, M´el´eard,...

Θ'N: Trait space

ni(t): Number of individuals of trait i∈Θ

Dynamics of the process

n(t) = (n0(t), n1(t), . . .)∈NN : Each individual of traiti

reproduces clonally with rate bi(1−µ)

reproduces with mutation with ratebiµaccording to some mutation kernel M(i, j)

dies due to age or competition with rate di+P

jcijnj

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Clonal reproduction model

Bolker, Pacala, Dieckmann, Law, Champagnat, M´el´eard,...

Θ'N: Trait space

ni(t): Number of individuals of trait i∈Θ

Dynamics of the process

n(t) = (n0(t), n1(t), . . .)∈NN : The populationni

increases by 1 with ratebi(1−µ)ni

makesnj increase by 1 with ratebiµ·M(i, j)·ni

deceases by 1 with rate(di+P

jcijnj)ni

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Clonal reproduction model

Bolker, Pacala, Dieckmann, Law, Champagnat, M´el´eard,...

Θ'N: Trait space

ni(t): Number of individuals of trait i∈Θ

Dynamics of the rescaledprocess (with competitioncij/K)

nK(t) = 1

K(n0(t), n1(t), . . .)∈(N/K)N : The populationnKi

increases by 1/K with ratebi(1−µ)·KnKi

makesnKj increase by1/K with ratebiµM(i, j)·KnKi deceases by 1/K with rate(di+P

jcijnKj )·KnKi

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Large populations limit

µ = 0 and K → ∞

Fournier, M´el´eard, 2004

Proposition

Let Θ ={1,2}, assume that the initial condition (nK1 (0), nK2 (0))

converges to a deterministic vector (x0, y0) for K → ∞. Then the process (nK1 (t), nK2 (t))converges in law, on bounded time intervals, to the

solution of d dt

x(t) y(t)

=X(x(t), y(t)) =

(b1d1c11xc12y)x (b2d2c21xc22y)y

. with initial condition (x(0), y(0)) = (x0, y0).

Monomorphic equilibria : n¯i= bic−di Invasion fitness : fij =bi−di−ciiij¯nj.

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Large populations and rare mutations limit

µ = µ(K) (K log K)

−1 Champagnat, 2006

Start with nKi (0) = ¯nx1i=x+K11i=y (time of the first mutation).

Ify is fitter than x, i.e. fyx>0 andfxy <0, then with proba→1:

O(logK) O(1) O(logK)

Supercritical BP LLN Subcritical BP 9 / 29

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Trait Substitution Sequence

Champagnat, 2006

Proposition Assume

logK 1 exp(cK)

∀(i, j)∈Θ2, either fji <0 orfji >0 andfij <0.

Then the rescaled process

(nKt/Kµ)t≥0 ⇒(¯nXt1Xt)t≥0 where Xt is a Markov chain onΘ with transition kernel

P(i, j) =bi[fji]+

bj M(i, j).

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Outline

1 Introduction

2 Clonal reproduction model

3 Mendelian diploid model Genetic variability

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Mendelian Diploid Model

Collet, M´el´eard, Metz, 2013

Let U be the allelic trait space, a countable set.

For example U ={a, A}.

An individualiis determined by two alleles out ofU :

I genotype: (ui1, ui2)∈ U2

I phenotype: φ((ui1, ui2)), withφ:U2R+

Rescaled population:

let Nt be the total number of individuals at timet,

nKu1,u2(t) = 1 K

Nt

X

i=1

1(ui

1,ui2)(t)

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Reproduction

Reproduction rate of (ui1, ui2) with (uj1, uj2):

fui

1ui2fuj

1uj2

Number of potential partners of(ui1, ui2)× their mean fertility

Reproduction without muta- tion: Probability1−µK

⇒ Mendelian rules: newborn get- ting genotype with coordinates that are sampled at random from each parent.

Ab aA bB

ab aB AB

Reproduction with mutation: Probability µK

⇒ changing one of the two allelic traits of the newborn from a to b according to the kernel M(a, b).

aA cC

âc Âc âC ÂC

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Reproduction

Reproduction rate of (ui1, ui2) with (uj1, uj2):

fui

1ui2fuj

1uj2

Number of potential partners of(ui1, ui2)× their mean fertility

Reproduction without muta- tion: Probability1−µK

⇒ Mendelian rules: newborn get- ting genotype with coordinates that are sampled at random from each parent.

Ab aA bB

ab aB AB

Reproduction with mutation: Probability µK

⇒ changing one of the two allelic traits of the newborn from a to b according to the kernel M(a, b).

aA cC

âc Âc âC ÂC

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Reproduction

Reproduction rate of (ui1, ui2) with (uj1, uj2):

fui

1ui2fuj

1uj2

Number of potential partners of(ui1, ui2)× their mean fertility

Reproduction without muta- tion: Probability1−µK

⇒ Mendelian rules: newborn get- ting genotype with coordinates that are sampled at random from each parent.

Ab aA bB

ab aB AB

Reproduction with mutation:

Probability µK

⇒ changing one of the two allelic traits of the newborn from a to b according to the kernel M(a, b).

aA cC

âc Âc âC ÂC

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Birth and Death Rate

Let U ={a, A}, andµK = 0.

The populations increase by 1 with rate

baa = (faanaa+12faAnaA)

2

faanaa+faAnaA+fAAnAA

baA = 2 (faanaaf+12faAnaA)(fAAnAA+12faAnaA)

aanaa+faAnaA+fAAnAA

bAA = (fAAnAA+12faAnaA)

2

faanaa+faAnaA+fAAnAA

The populations decrease by 1 with rate

daa = (Daa+Caa,aanaa+Caa,aAnaA+Caa,AAnAA)naa

daA = (DaA+CaA,aanaa+CaA,aAnaA+CaA,AAnAA)naA dAA = (DAA+CAA,AAnAA+CAA,aAnaA+CAA,aanaa)nAA

aa

aa aa aa aA aA aa aA aA

1

1/4 1/2

1/2

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Birth and Death Rate

Let U ={a, A}, andµK = 0.

The populations increase by 1 with rate

baa = (faanaa+12faAnaA)

2

faanaa+faAnaA+fAAnAA

baA = 2 (faanaaf+12faAnaA)(fAAnAA+12faAnaA)

aanaa+faAnaA+fAAnAA

bAA = (fAAnAA+12faAnaA)

2

faanaa+faAnaA+fAAnAA

The populations decrease by 1 with rate

daa = (Daa+Caa,aanaa+Caa,aAnaA+Caa,AAnAA)naa

daA = (DaA+CaA,aanaa+CaA,aAnaA+CaA,AAnAA)naA dAA = (DAA+CAA,AAnAA+CAA,aAnaA+CAA,aanaa)nAA

aa

aa aa aa aA aA aa aA aA

1

1/4 1/2

1/2

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Large populations limit

Collet, M´el´eard, Metz, 2013

Proposition

Assume that the initial condition (nKaa(0), nKaA(0), nKAA(0)) converges to a deterministic vector (x0, y0, z0) for K→ ∞. Then the process

(nKaa(t), nKaA(t), nKAA(t))converges in law to the solution of

d dt

x(t) y(t) z(t)

!

=X(x(t), y(t), z(t)) =

baa(x, y, z)daa(x, y, z) baA(x, y, z)daA(x, y, z) bAA(x, y, z)dAA(x, y, z)

! .

bAA(x, y, z) = (fAAz+

1 2faAy)2 faax+faAy+fAAz

dAA(x, y, z) = (DAA+CAA,aax+CAA,aAy+CAA,AAz)z and similar expressions for the other terms

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The 3-System (aa, aA, AA)

Phenotypic viewpoint allele A dominant allele a recessive

The dominant allele A defines the phenotype⇒ φ(aA) =φ(AA)

cu1u2,v1v2 ≡c, ∀u1u2, v1v2 ∈ {aa, aA, AA} fAA=faA=faa≡f

DAA=DaA≡DbutDaa =D+∆

⇒ type aA is as fit asAA and both are fitter than type aa

AA aA

aa

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The 3-System (aa, aA, AA)

Phenotypic viewpoint allele A dominant allele a recessive

The dominant allele A defines the phenotype⇒ φ(aA) =φ(AA)

cu1u2,v1v2 ≡c, ∀u1u2, v1v2 ∈ {aa, aA, AA}

fAA=faA=faa≡f

DAA=DaA≡DbutDaa =D+∆

⇒ type aA is as fit asAA and both are fitter than type aa

AA aA

aa

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The 3-System (aa, aA, AA)

Phenotypic viewpoint allele A dominant allele a recessive

The dominant allele A defines the phenotype⇒ φ(aA) =φ(AA)

cu1u2,v1v2 ≡c, ∀u1u2, v1v2 ∈ {aa, aA, AA}

fAA=faA=faa≡f

DAA=DaA≡DbutDaa =D+∆

⇒ type aA is as fit asAA and both are fitter than type aa

AA aA

aa

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Work of Bovier, Neukirch (2015)

Start with nKi (t) = ¯naa1i=aa+K11i=aA then :

O(logK) O(1) ≥O(K1/4+)

The system converges to(0,0,n¯AA) ast→ ∞ butslowly!

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Genetic Variability ? Polymorphism?

Suppose a new dominant mutant allele B appears before aA dies out.

Suppose that phenotypes a and B cannot reproduce.

Can the aa-population recover and coexist with the mutant population ?

We will study the deterministic sytem and start with initial condition ni(0) = ¯nAA1i=AA+1i=aA+21i=aa+31i=AB

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Model with a second mutant

Mutation to allele B → U ={a, A,B}

aa aA AA aB AB BB

⇒ 6 possible genotypes: aa,aA,AA,aB,AB,BB

⇒ Dominance of allelesa < A < B Differences in Fitness:

fertility: fa=fA=fB=f

natural death: Da=D+ ∆> DA=D > DB=D−∆

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Model with a second mutant

Mutation to allele B → U ={a, A,B}

aa aA AA aB AB BB

⇒ 6 possible genotypes: aa,aA,AA,aB,AB,BB

⇒ Dominance of allelesa < A < B

Differences in Fitness:

fertility: fa=fA=fB=f

natural death: Da=D+ ∆> DA=D > DB=D−∆

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Model with a second mutant

Mutation to allele B → U ={a, A,B}

aa aA AA aB AB BB

⇒ 6 possible genotypes: aa,aA,AA,aB,AB,BB

⇒ Dominance of allelesa < A < B Differences in Fitness:

fertility: fa=fA=fB=f

natural death: Da=D+ ∆> DA=D > DB=D−∆

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Birth Rates

No recombination between a and B

BB a

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Birth Rates

birth-rate of aa-individual:

baa=naa naa+12naA

Pool(aa) +

1 2naB 1

2naA+12naB

Pool(aB)

+

1

2naA naa+12naA+ 12naB Pool(aA)

Pools of potential partners:

aa aA aB

aa aA aa aA aA

AA AA aB AB AA aB AB

BB BB

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Competition

No competition between a and B

BB a

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Competition - first try

aa aA AA aB AB BB

aa c c c 0 0 0

aA c c c c c c

AA c c c c c c

aB 0 c c c c c

AB 0 c c c c c

BB 0 c c c c c

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Competition - first try

0 100 200 300 400 500 600 700

1 2 3 4 5

aaaA AAaBABBB 24 / 29

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Competition - first try

100 200 300 400 500 600 700

-35 -30 -25 -20 -15 -10 -5

aaaA AAaBABBB

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Competition - second try

competition felt by naA fromB-individuals

< competition felt by nAA fromB-individuals

the decay ofaA-population slows down aa-population can recover

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Competition - second try

aa aA AA aB AB BB

aa c c c 0 0 0

aA c c c c c c−η

AA c c c c c c

aB 0 c c c c c

AB 0 c c c c c

BB 0 c−η c c c c

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Competition - second try

0 100 200 300 400 500 600 700

1 2 3 4 5

aaaA AAaBABBB 27 / 29

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Competition - second try

100 200 300 400 500 600 700

-35 -30 -25 -20 -15 -10 -5

aaaA AAaBABBB

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Recap - Dimorphism in two mutations

Mutation 1 Mutation 2

0 100 200 300 400 500 600 700

1 2 3 4 5

0 100 200 300 400 500 600 700

1 2 3 4 5

100 200 300 400 500 600 700

-35 -30 -25 -20 -15 -10 -5

100 200 300 400 500 600 700

-35 -30 -25 -20 -15 -10 -5

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Proof

1.Phase: Fixation of the mutant

AB grows to levelε0

aB, BB≤ε0

⇒ perturbation of the 3-system(aa, aA, AA) of at mostO(ε0)

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Proof

2.Phase: Invasion of the mutant

aa, aA, aB≤ε0

⇒ perturbation of the new 3-system (AA, AB, BB) of at most O(ε0)

⇒ use results of Bovier, Neukirch (2015) naA+naB increases ifη >0

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Proof

3. Phase: Recovery ofaa

AAsmall enough aAbig enough

⇒ aastarts to reproduce out of itself as much as with the other partners.

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Proof

4. Phase: Coexistence

Delicate phase : aagrows out of itself and feels no competition with BB

⇒ convergence to coexistence-fixed-point n¯aa,BB BUT meanwhile :

due to Mendelian recombination,aA, aB, AB have a ”bump” upwards, and due to competition with them BB has a ”bump” downwards.

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T H A N K

Y O U

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