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 P 1 = phage strain P 2 = another phage strain L 1 = lysogens of phage P 1 L 2 = lysogens of phage P 2 The only differences between P 1 and P 2 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 P 1 is the resident phage. Assume that the system has reached its equilibrium (R*, S*, L 1 *, P 1 *) Can P 2 invade?

Linearization around the equilibrium To see if P 2 can invade, consider the linearized system: P 2 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 (p 1 =p 2 =p*), the first order partial derivatives of the fitness function with respect to p 1 and p 2 will be equal to zero Differentiating s p1 (p 2 ) 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 p 2 Let b be the second order derivative of the fitness function with respect to p 2, evaluated at the singularity p*: Then

Differentiating with respect to p 2 Let a be the second order derivative of the fitness function with respect to p 1, 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.