Evolution of Asymmetric Damage Segregation : A Modelling Approach

Abstract

Mother cell-specific ageing is a well-known phenomenon in budding yeast Saccharomyces cerevisiae. Asymmetric segregation of damage and its accumulation in the mother cell has been proposed as one important mechanism. There are, however, unicellular organisms such as the fission yeast Schizosaccharomyces pombe, which replicates with almost no asymmetry of segregation of damage and the pathogenic yeast Candida albicans, which falls around the middle of the segregation spectrum far from both complete symmetry and complete asymmetry . The ultimate evolutionary cause that determines the way damage segregates in a given organism is not known. Here we develop a mathematical model to examine the selective forces that drive the evolution of asymmetry and discover the conditions in which symmetry is the optimal strategy. Three main processes are included in the model: protein synthesis (growth), protein damage , and degradation of damage . We consider, for the first time, the costs to the cell that might accompany the evolution of asymmetry and incorporate them into the model along with known trade-offs between reproductive and maintenance investments and their energy requirements. The model provides insight into the relationship between ecology and cellular trade-off physiology in the context of unicellular ageing , and applications of the model may extend to multicellular organisms.

Appendix

Here we show that with conditions introduced in equation (14.3) in the text, all cells survive under symmetry, that is to say, symmetry is the best strategy. By looking at Fig. 14.1b, we realize that the condition for survival of the cell is

$$D_T < D^*$$

Considering the equation for D _T derived in section “With Degradation”, we will have

$$D\left(0\right) < s\Delta /g\left(\Delta - g\right) + \left[D^* + \Delta \left(g - s\right)/g\left(\Delta - g\right)\right]\left[s/\left(s - \Delta \right)\right]^{g/\Delta }$$

If the right-hand side of the above equation is larger than D ^*, it can easily be seen that everybody will survive. For this to happen we need

$$\left[s\Delta /g\left(\Delta - g\right) - D^* \right]\left\{ 1 - \left[s/\left(s - \Delta \right)\right]^{g/\Delta } \right\} + \left[\Delta /\left(\Delta - g\right)\right]\left[s/\left(s - \Delta \right)\right]^{g/\Delta } > 0,$$

which then gives

$$s\Delta /g\left(\Delta - g\right) - D^* < \left[\Delta /\left(\Delta - g\right)\right]\left[s/\left(s - \Delta \right)\right]^{g/\Delta } \left\{ \left[s/\left(s - \Delta \right)\right]^{g/\Delta } - 1\right\} ^{ - 1} $$

((14.7))

The right-hand side of the above equation is larger than 1 for g < Δ. For (7) to hold, we now only require

$$s\Delta /g\left(\Delta - g\right) - D^* < 1$$

Equivalently, we need

$$g^2 - \Delta g + s\Delta /\left(1 + D^* \right) < 0$$

This requires (for the above equation to have real roots)

$$s < \Delta \left(1 + D^* \right)/4$$

((14.8))

and

$$\left(\Delta - \sqrt {\Delta ^2 - 4s\Delta /\left(1 + D^* \right)} \right)/2 < g < \left(\Delta + \sqrt {\Delta ^2 - 4s\Delta /\left(1 + D^* \right)} \right)/2$$

((14.9))

Putting (14.8) and (14.9) together with g < Δ, we have the conditions derived in (14.3). Note that these are sufficient, but not necessary, conditions.

Note

While this chapter was under review, an interesting article was published by Lindner et al. (2008). The authors demonstrated asymmetric segregation of protein aggregates between the offspring in Escherichia coli. Accumulation of the aggregates to the older pole of the cell results in a progressively ageing mother cell (i.e. old-pole progeny) and rejuvenated new-pole progeny. The authors showed that the segregation of protein aggregation is associated with significant loss of reproductive ability in the old-pole progeny compared to the new-pole progeny. Also, Klinger et al. (2010) showed in another article that the oxidatively inactivated acotinase in budding yeast is distributed between the mother cell and the daughter cell according to volume asymmetry , but the still active part of the enzyme is preferentially segregated to the daughter cell. The authors suggested that this process aids the rejuvenation of the daughter cell.