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.
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