Adaptive Dynamics of Temperate Phages

Introduction

• Phages are viruses which infect bacteria• A temperate phage can either replicate lytically or

lysogenically• Lysis means the phage makes many copies of itself

and releases the new phages by bursting the bacteria open. Bacteria is destroyed.

• Lysogeny means the phage inserts its DNA into the bacterial DNA and is replicated passively when the bacteria divides. Bacteria (lysogen) survives.

• Lysogens can later be induced, i.e. phage DNA extricates itself from the bacterial DNA and carries out lysis.

Lysis looks like this

The populations in the model

• R = resources• S = sensitive bacteria

• P1 = phage strain

• P2 = another phage strain

• L1 = lysogens of phage P1

• L2 = lysogens of phage P2

• The only differences between P1 and P2 are that they have different probabilities of lysogeny and different induction rates.

Some important parameters

• ω = chemostat flow rate

• δ = adsorption rate

• p = probability of lysogeny

• (1-p) = probability of lysis

• i = induction rate

• β = burst size

The Model

Invasion of resident strain by a mutant

• Suppose P1 is the resident phage.

• Assume that the system has reached its equilibrium (R*, S*, L1*, P1*)

• Can P2 invade?

Linearization around the equilibrium

• To see if P2 can invade, consider the linearized system:

• P2 can invade if there is a positive eigenvalue

The fitness function

• It turns out that there will be at least one positive eigenvalue as long as the following condition is satisfied:

Q, μ, and γ

Introducing a trade-off function

• Now let i = f(p)

• Fitness function becomes:p

i

Evolutionary singularities

• At an evolutionary singularity (p1=p2=p*), the first order partial derivatives of the fitness function with respect to p1 and p2 will be equal to zero

• Differentiating sp1(p2) with respect to p2 and setting equal to zero:

• So at a singularity p*, we must have either or

Identifying evolutionary singularities

Branching points

Evolutionary branching

• Let p* be an evolutionary singularity• Let

• Then p* will be a branching point if (i) b>0 (i.e. p* is not ESS) (ii) (a-b)>0 (i.e. p* is CS)

Differentiating with respect to p2

• Let b be the second order derivative of the fitness function with respect to p2,

evaluated at the singularity p*:

• Then

Differentiating with respect to p2

• Let a be the second order derivative of the fitness function with respect to p1,

evaluated at the singularity p*.

• It turns out that:

• Suppose b>0 (i.e. singularity is not ESS)

• For evolutionary branching, we also need (a-b)>0 (i.e. singularity is CS).

• From previous slide:

• So we need to find the derivative of μ at the singularity

Finding the derivative of μ

• Start from the resident ODEs at equilibrium:

Derivative of μ is zero

• Remember that μ(p)=δS(p)P(p)/L(p)• We know the derivatives of S, P and L are all zero • So by the quotient rule, the derivative of μ must also

be zero.• So from

we find that i.e. branching is not possible.