Selective advantage for sexual reproduction with random haploid fusion

Abstract

This article develops a simplified set of models describing asexual and sexual replication in unicellular diploid organisms. The models assume organisms whose genomes consist of two chromosomes, where each chromosome is assumed to be functional if it is equal to some master sequence σ₀, and non-functional otherwise. We review the previously studied case of selective mating, where it is assumed that only haploids with functional chromosomes can fuse, and also consider the case of random haploid fusion. When the cost for sex is small, as measured by the ratio of the characteristic haploid fusion time to the characteristic growth time, we find that sexual replication with random haploid fusion leads to a greater mean fitness for the population than a purely asexual strategy. However, independently of the cost for sex, we find that sexual replication with a selective mating strategy leads to a higher mean fitness than the random mating strategy. The results of this article are consistent with previous studies suggesting that sex is favored at intermediate mutation rates, for slowly replicating organisms, and at high population densities. Furthermore, the results of this article provide a basis for understanding sex as a stress response in unicellular organisms such as Saccharomyces cerevisiae (Baker’s yeast).

Appendices

Appendix 1: Derivation of the steady-state mean fitness for the selective mating strategy

For the selective mating strategy, the steady-state equations are given by,

$$ \begin{aligned} &0 = -(\kappa_{\rm vv} + \bar{\kappa}(t = \infty)) x_{\rm vv} + \gamma \rho^{*} x_{\rm v}^2 A_{\rm vv} \\ & 0 = -(\kappa_{\rm vu} + \bar{\kappa}(t = \infty)) x_{\rm vu} + \gamma \rho^{*} x_{\rm v}^2 B_{\rm vv} \\ & 0 = -\bar{\kappa}(t = \infty) x_{\rm v} + 2 \kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu} - \gamma \rho^{*} x_{\rm v}^2 \\ & 0 = -\bar{\kappa}(t = \infty) x_{\rm u} + \kappa_{\rm vu} x_{\rm vu} \end{aligned} $$

(A1)

where $\bar{\kappa}(t = \infty) = -\kappa_{\rm vv} x_{\rm vv} - \kappa_{\rm vu} x_{\rm vu} + \gamma \rho^{*} x_{\rm v}^2 .$ We purposely neglect the steady-state equation corresponding to d x _uu/dt = 0, since this equation will not be necessary to determine the mean fitness of the population.

Solving the last equation for γρ* x ²_v and substituting the result into the expression for $\bar{\kappa}(t = \infty)$ gives,

$$ \bar{\kappa}(t = \infty) = \kappa_{\rm vv} x_{\rm vv} - \bar{\kappa}(t = \infty) x_{\rm v} $$

(A2)

and so,

$$ x_{\rm v} = {\frac{\kappa_{\rm vv} x_{\rm vv}}{\bar{\kappa}(t = \infty)}} - 1 $$

(A3)

We also have,

$$ x_{\rm u} = {\frac{\kappa_{\rm vu} x_{\rm vu}}{\bar{\kappa}(t = \infty)}} $$

(A4)

Substituting these values into the expression for $\bar{\kappa}(t = \infty) ,$ and making use of the fact that ρ* = ρ/(1 + (1/2) (x _v + x _u)) gives, after some manipulation,

$$ {\frac{(x_{\rm vv} - \phi_{\rm ss})^2}{\phi_{\rm ss} (x_{\rm vv} + \alpha x_{\rm vu} + \phi_{\rm ss})^2}} = {\frac{1}{2}} {\frac{\kappa_{\rm vv}}{\gamma \rho}} $$

(A5)

The steady-state equations for x _vv and x _vu then give,

$$ \begin{aligned} & (1 + \phi_{\rm ss}) x_{\rm vv} = A_{\rm vv} (x_{\rm vv} + \alpha x_{\rm vu} + \phi_{\rm ss}) \\ & (\alpha + \phi_{\rm ss}) x_{\rm vu} = B_{\rm vv} (x_{\rm vv} + \alpha x_{\rm vu} + \phi_{\rm ss}) \end{aligned} $$

(A6)

Therefore, x _vu/x _vv = B _vv/A _vv × (1 + ϕ_ss)/(α + ϕ_ss), so that

$$ (1 + \phi_{\rm ss}) x_{\rm vv} = A_{\rm vv} \phi_{\rm ss} + \left(A_{\rm vv} + (1 + \phi_{\rm ss}) B_{\rm vv} {\frac{\alpha}{\alpha + \phi_{\rm ss}}}\right) x_{\rm vv} \Rightarrow x_{\rm vv} = {\frac{A_{\rm vv} \phi_{\rm ss} (\alpha + \phi_{\rm ss})} {(1 + \phi_{\rm ss}) (\alpha + \phi_{\rm ss} - B_{\rm vv} \alpha) - A_{\rm vv} (\alpha + \phi_{\rm ss})}} $$

(A7)

Now, the first steady-state equation for x _vv may be solved to give, x _vv + αx _vu + ϕ_ss = (1 + ϕ_ss) x _vv/A _vv, which may be substituted into Eq. A5 to give,

$$ {\frac{A_{\rm vv}^2 \left(1 - {\frac{\phi_{\rm ss}}{x_{\rm vv}}}\right)^2}{\phi_{\rm ss} (1 + \phi_{\rm ss})^2}} = {\frac{1}{2}} {\frac{\kappa_{\rm vv}}{\gamma \rho}} $$

(A8)

Substituting in the value for x _vv gives, after some manipulation, Eq. 6 defining the steady-state mean fitness.

Appendix 2: Derivation of the steady-state mean fitness for the random mating strategy

For the random mating strategy, the steady-state equations are,

$$ \begin{aligned} 0 &= -(\kappa_{\rm vv} + \bar{\kappa}(t = \infty)) x_{\rm vv} + \gamma \rho^{*} x_{\rm v}^2 A_{\rm vv} \\ 0 &= -(\kappa_{\rm vu} + \bar{\kappa}(t = \infty)) x_{\rm vu} + \gamma \rho^{*} x_{\rm v}^2 B_{\rm vv} + \gamma \rho^{*} x_{\rm v} x_{\rm u} B_{\rm vu} \\ 0 &= -\bar{\kappa}(t = \infty) (x_{\rm v} + x_{\rm u}) + 2 (\kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu}) \\ & \quad- \gamma \rho^{*} (x_{\rm v} + x_{\rm u})^2 \\ 0 &= -\bar{\kappa}(t = \infty) x_{\rm v} + 2 \kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu} - \gamma \rho^{*} x_{\rm v} (x_{\rm v} + x_{\rm u}) \end{aligned} $$

(B1)

where $\bar{\kappa}(t = \infty) = -\kappa_{\rm vv} x_{\rm vv} - \kappa_{\rm vu} x_{\rm vu} + \gamma \rho^{*} (x_{\rm v} + x_{\rm u})^2 .$ The third equation is obtained by adding the equations corresponding to d x _v/dt = d x _u/dt = 0.

We then have,

$$ \bar{\kappa}(t = \infty) = \kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu} - \bar{\kappa}(t = \infty) (x_{\rm v} + x_{\rm u}) $$

(B2)

so that,

$$ x_{\rm v} + x_{\rm u} = {\frac{\kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu}} {\bar{\kappa}(t = \infty)}} - 1 $$

(B3)

If we define $\tilde{x}_{\rm v} = x_{\rm v}/(x_{\rm v} + x_{\rm u}) ,\; \tilde{x}_{\rm u} = x_{\rm u}/(x_{\rm v} + x_{\rm u}) ,$ then,

$$ 0 = 2 \kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu} - 2 \tilde{x}_{\rm v} (\kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu}) $$

(B4)

and so,

$$ \begin{aligned} & \tilde{x}_{\rm v} = {\frac{2 \kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu}}{2 (\kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu})}}\\ & \tilde{x}_{\rm u} = {\frac{\kappa_{\rm vu} x_{\rm vu}}{2 (\kappa_{\rm vv} x_{\rm vv} + \kappa_{\rm vu} x_{\rm vu})}} \end{aligned} $$

(B5)

By substituting the expression for x _v + x _u into the expression defining $\bar{\kappa}(t = \infty) ,$ we obtain, after some manipulation,

$$ {\frac{(x_{\rm vv} + \alpha x_{\rm vu} - \phi_{\rm rs})^2}{\phi_{\rm rs} (x_{\rm vv} + \alpha x_{\rm vu} + \phi_{\rm rs})^2}} = {\frac{1}{2}} {\frac{\kappa_{\rm vv}} {\gamma \rho}} $$

(B6)

The steady-state equations for x _vv and x _vu give,

$$ \begin{aligned} (1 + \phi_{\rm rs}) x_{\rm vv} =& \,A_{\rm vv} \left({\frac{2 x_{\rm vv} + \alpha x_{\rm vu}}{2 (x_{\rm vv} + \alpha x_{\rm vu})}}\right)^2 (\phi_{\rm rs} + x_{\rm vv} + \alpha x_{\rm vu}) \\ (\alpha + \phi_{\rm rs}) x_{\rm vu} =& \,B_{\rm vv} \left({\frac{2 x_{\rm vv} + \alpha x_{\rm vu}}{2 (x_{\rm vv} + \alpha x_{\rm vu})}}\right)^2 (\phi_{\rm rs} + x_{\rm vv} + \alpha x_{\rm vu}) \\ & + B_{\rm vu} {\frac{(2 x_{\rm vv} + \alpha x_{\rm vu}) \alpha x_{\rm vu}}{4 (x_{\rm vv} + \alpha x_{\rm vu})^2}} (\phi_{\rm rs} + x_{\rm vv} + \alpha x_{\rm vu}) \end{aligned} $$

(B7)

Now, from Eq. B2 it can be seen that ϕ_rs ≤ x _vv + α x _vu, and so from Eq. B6 we have,

$$ {\frac{x_{\rm vv} + \alpha x_{\rm vu} - \phi_{\rm rs}}{x_{\rm vv} + \alpha x_{\rm vu} + \phi_{\rm rs}}} = \sqrt{{\frac{\kappa_{\rm vv}}{2 \gamma \rho}} \phi_{\rm rs}} \Rightarrow x_{\rm vv} + \alpha x_{\rm vu} = \phi_{\rm rs} f(\phi_{\rm rs}, \lambda) $$

(B8)

where λ ≡ κ_vv/(2γρ), $f(\phi_{\rm rs}, \lambda) \equiv (1 + \sqrt{\lambda \phi_{\rm rs}})/(1 - \sqrt{\lambda \phi_{\rm rs}}) .$

We now have,

$$ (1 + \phi_{\rm rs}) x_{\rm vv} = A_{\rm vv} (1 + f(\phi_{\rm rs}, \lambda)) {\frac{(x_{\rm vv} + \phi_{\rm rs} f(\phi_{\rm rs}, \lambda))^2}{4 \phi_{\rm rs} f(\phi_{\rm rs}, \lambda)^2}} (\alpha + \phi_{\rm rs}) {\frac{\phi_{\rm rs} f(\phi_{\rm rs}, \lambda) - x_{\rm vv}} {\alpha}} = B_{\rm vv} (1 + f(\phi_{\rm rs}, \lambda)) \times{\frac{(x_{\rm vv} + \phi_{\rm rs} f(\phi_{\rm rs}, \lambda))^2}{4 \phi_{\rm rs} f(\phi_{\rm rs}, \lambda)^2}} + B_{\rm vu} (1 + f(\phi_{\rm rs}, \lambda)) \times {\frac{(\phi_{\rm rs} f(\phi_{\rm rs}, \lambda) + x_{\rm vv}) (\phi_{\rm rs} f(\phi_{\rm rs}, \lambda) - x_{\rm vv})}{4 \phi_{\rm rs} f(\phi_{\rm rs}, \lambda)^2}} $$

(B9)

The second equation gives,

$$ (\alpha + \phi_{\rm rs}) \phi_{\rm rs} f(\phi_{\rm rs}, \lambda) - (\alpha + \phi_{\rm rs}) x_{\rm vv} = \alpha {\frac{1 + f(\phi_{\rm rs}, \lambda)}{4 \phi_{\rm rs} f(\phi_{\rm rs}, \lambda)^2}} \times [B_{\rm vv} (\phi_{\rm rs} f(\phi_{\rm rs}, \lambda) + x_{\rm vv})^2 + B_{\rm vu} (\phi_{\rm rs}^2 f(\phi_{\rm rs}, \lambda)^2 - x_{\rm vv}^2)] $$

(B10)

Now, ϕ ²_rs f(ϕ_rs, λ)² − x ²_vv  = −(ϕ_rs f(ϕ_rs, λ) + x _vv)² + 2ϕ ²_rs f(ϕ_rs, λ)² + 2ϕ_rs f(ϕ_rs, λ) x _vv, and so, after some manipulation, we obtain,

$$ x_{{{\text{vv}}}} = \phi _{{{\text{rs}}}} f(\phi _{{{\text{rs}}}} ,\lambda ) \times {\left[ {(\alpha + \phi _{{{\text{rs}}}} )f(\phi _{{{\text{rs}}}} ,\lambda ) - \frac{1}{2}\alpha B_{{{\text{vu}}}} (1 + f(\phi _{{{\text{rs}}}} ,\lambda ))} \right]} \mathord{\left/ {\vphantom {{\left[ {(\alpha + \phi _{{{\text{rs}}}} )f(\phi _{{{\text{rs}}}} ,\lambda ) - \frac{1}{2}\alpha B_{{{\text{vu}}}} (1 + f(\phi _{{{\text{rs}}}} ,\lambda ))} \right]}}} \right. \kern-\nulldelimiterspace} \left[\vphantom{\frac{1}{2}} \phi _{{{\text{rs}}}} f(\phi _{{{\text{rs}}}} ,\lambda )(1 - 2\alpha )\right.\left. + \alpha \left( {\frac{1}{2}B_{{{\text{vu}}}} (1 + f(\phi _{{{\text{rs}}}} ,\lambda )) - f(\phi _{{{\text{rs}}}} ,\lambda )} \right) \right] $$

(B11)

where we made use of the fact that B _vv − B _vu = −2A _vv and 2(1 + ϕ_rs) x _vv = (1 + f(ϕ_rs, λ))/(2ϕ_rs f(ϕ_rs, λ)²) A _vv (ϕ_rs f(ϕ_rs, λ) + x _vv)².

Plugging the value of x _vv back into the first equation from Eq. B9 we obtain, after tedious algebra,

$$ [1 + \phi_{\rm rs}] \times \left[\phi_{\rm rs} f(\phi_{\rm rs}, \lambda) - \alpha\left({\frac{1}{2}} B_{\rm vu} (1 + f(\phi_{\rm rs}, \lambda)) - f(\phi_{\rm rs}, \lambda)\right)\right] \times \left[\phi_{\rm rs} f(\phi_{\rm rs}, \lambda) (1 - 2 \alpha) + \alpha \left({\frac{1}{2}} B_{\rm vu} (1 + f(\phi_{\rm rs}, \lambda)) - f(\phi_{\rm rs}, \lambda)\right)\right] = A_{\rm vv} (1 - \alpha)^2 f(\phi_{\rm rs}, \lambda) (1 + f(\phi_{\rm rs}, \lambda)) \phi_{\rm rs}^2 $$

(B12)

Now, $1 + f(\phi_{\rm rs}, \lambda) = 2/(1 - \sqrt{\lambda \phi_{\rm rs}}) ,$ and so, multiplying both sides by $(1 - \sqrt{\lambda \phi_{\rm rs}})^2$ gives Eq. (7) defining the steady-state mean fitness.