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Adaptive Dynamics of Temperate Phages

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.

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Adaptive Dynamics of Temperate Phages

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  1. Adaptive Dynamics of Temperate Phages

  2. 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.

  3. Lysis looks like this

  4. 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.

  5. 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

  6. The Model

  7. 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?

  8. Linearization around the equilibrium • To see if P2 can invade, consider the linearized system: • P2 can invade if there is a positive eigenvalue

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

  10. Q, μ, and γ

  11. Introducing a trade-off function • Now let i = f(p) • Fitness function becomes: i p

  12. 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

  13. Identifying evolutionary singularities

  14. Branching points

  15. 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)

  16. 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

  17. 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:

  18. 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

  19. Finding the derivative of μ • Start from the resident ODEs at equilibrium:

  20. 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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