Lecture of April 17, 2018: Evolutionary Dynamics: Replication, Selection and Mutation

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Lecture of April 17, 2018: Evolutionary Dynamics:

Replication, Selection and Mutation

Chiara Mocenni

Course on Game Theory

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Replication

Let

˙ x = rx ,

where x (t) is the abundance of cells at time t. Suppose that cells replicate at a rate r . The solution of this equation is

x (t) = x

0

e

^rt

.

For r > 0 the cells grow to infinity For r < 0 the cells decrease to 0

For r = 0 the cell number do not change respect to the initial

value x

₀

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Replication

If we also have a mortality term with mortality rate d , the equation becomes:

˙

x = (r − d )x .

For r > d the cells grow to infinity For r < d the cells decrease to 0

For r = d the cell number do not change respect to the initial

value x

₀

.

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Selection

Suppose to have two types of individuals that reproduce with different rates. A replicates with reproduction rate a and B replicates with reproduction rate b. The reproduction rate is the fitness of the type.

Let x (t) the number of individuals of type A and y (t) the number of individuals of type B, and let x (0) = x

₀

and y (0) = y

₀

the initial distribution of the two types in the population. The two subpopulations evolve as ˙ x = ax and ˙ y = by , and x (t) = x

0

e

^at

, y (t) = x

₀

e

^bt

.

x (t + τ ) = 2x (t) implies that x

₀

e

^{a(t+τ )}

= 2x

₀

e

^at

, then

e

^at

e

^aτ

= 2e

^at

, e

^aτ

= 2. Thus, τ =

^{log 2}_a

. Analogously, for type B we have that τ =

^{log 2}_b

. In conclusion, if a > b then A is replicating in a shorter time compared to b and after some time in the

population there will be more individuals of type A then individuals

of type B.

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Mutation

Let u

1

and u

2

be the mutation rates from type A to type B and form type B to type A, respectively. The equations reads:

˙

x = x (1 − u

1

) + u

2

y − φx ,

˙

y = y (1 − u

1

) + u

1

x − φy ,

If a = b = 1 then φ = x + y = 1. Then the equation becomes

˙

x = u

₂

− (u

₁

+ u

₂

)x . The equation has one stable steady state

x

^∗

= u

2

u

1

+ u

2

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Selection

If we assume that the number of individuals is constant, e.e.

suppose that

x + y = 1,

we have that the variables x and y are frequencies and the two equations that regulate the dynamics of the two types will be

˙

x = x (a − φ),

˙

y = y (b − φ),

where φ is taking into account the constraint x + y = 1. This is true when

φ = ax + by .

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Selection

From x + y = 1 we have that

y = 1 − x ,

from which we have that ˙ y = − ˙ x and by means of the equation for y we can reduce the two equations into one as follows:

− ˙x = (1−x)(b−φ) = (1−x)(b−ax−by ) = (1−x)(b−ax−b(1−x)) =

= (1 − x )(−ax + bx ) = x (1 − x )(−a + b).

From which it follows that

˙

x = x (1 − x )(a − b).

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Selection

The equation has two steady states:

x

1

= 0: all individuals of type B, x

₂

= 1: all individuals of type A.

For a > b we have that ˙ x > 0 for 0 < x < 1, then x (t) → 1 and y (t) → 0.

For a < b we have that ˙ x < 0 for 0 < x < 1, then x (t) → 0

and y (t) → 1.