Fig. 7.6 Pathways toward the selection of genetic instability and cancer progression.

confers genetic instability. This mutant will be selected for instantly. (2) In the second pathway, the initial mutation confers genetic instability to the cell. Since apoptotic responses are still intact, the model analysis tells us that this variant will have a lower fitness compared to the wild type and will be on its way to extinction. However, because the cell is unstable, it can generate mutations at a faster rate. Thus, there is a chance that the mutation impairing apoptosis is generated before this cell variant has gone extinct. As soon as the apoptotic mechanism has been impaired, the unstable cell gains a selective advantage.

We can calculate which of these two pathways occurs faster, and this is the pathway that is more likely to lead to selection of instability (for mathematical details of this approach, see Chapters 2 and 4). We introduce the following notation (Figure 7.6). The number of cells in a tissue is denoted by N. The rate at which a genetically stable cell mutates (to be either unstable or apoptosis impaired) is given by fi. The rate at which an unstable cell mutates toward a loss of apoptotic function is denoted by fi. Thus, p, > p.. The relative reproductive rate of an unstable cell which has intact apoptotic responses is given by a < 1 (we assume that the wild-type reproductive rate is 1), which reflects the fact that unstable cells with intact apoptosis have a selective disadvantage. The rate at which an advantageous mutator is generated via the first pathway (first apoptosis~, then repair-) is given by p?N. The rate at which an advantageous mutator is generated via the second pathway (first repair~, then apoptosis~) is given by N/iaii/(l — a). Therefore, if a then repair and stability are lost first. In the opposite case, apoptosis is lost first. In biological terms, if the relative fitness of the unstable and apoptosis competent cell is sufficiently low (because mutants are killed efficiently) , then generation of an advantageous mutator is likely to proceed by first losing apoptosis, and then acquiring genetic instability. Knowledge of parameter values will be required to determine which of the two pathways is more likely. The result might vary between different tissues.

7.6 Can competition be reversed by chemotherapy?

The results derived in this chapter have implications for the use of chemotherapy (Figure 7.7). Chemotherapy essentially increases the degree of DNA damage. Therefore, it can be used to reverse the relative fitness of stable and unstable cells such that unstable cells are excluded (Figure 7.7). This can drive progressing cancer cells extinct and result in the persistence of stable cells. These may either be healthy cells or less aggressive and slowly progressing tumor cells. In order to achieve this outcome, there needs to be a sufficient difference in the repair rate between stable and unstable cells. If this is not the case, therapy can merely slow down cancer progression.

Since in this scenario, chemotherapy acts by modulating the competition between stable and unstable cells, it is not a requirement that every last cell is killed by the drugs. Selection and competition will make sure that the unstable cancer cells are driven extinct. This argument, however, requires that there is an element of competition between unstable and stable cells. Whether and under which circumstances this is the case remains to be determined.

This is a different mechanism of action compared to the traditional view which assumes that chemotherapy only acts by killing dividing cells. For chemotherapy to reverse the fitness of stable and unstable cells, two conditions are required, (i) There needs to be a sufficient difference in the repair rate between stable and unstable cells. The higher the replication rate of unstable cells relative to stable cells, the higher this difference in the repair rate required to achieve success. Therefore, contrary to traditional views, a faster rate of cell division of cancerous unstable cells renders successful

unstable cells

(a) Large difference in repair rate

(b) Small difference in repair rate

0 50Q 1000 1500

Time (arbitrary scale)

Fig. 7.7 Simulation of chemotherapy, modeled by an increase in the DNA hit rate, u. We start with a situation where cells which are unstable and have impaired apoptosis spontaneously give rise to cancer growth and progression, (a) If there is a large difference in repair rate between stable and unstable cells, therapy can exclude the unstable cells, (b) If there is a smaller difference between stable and unstable cells, then therapy fails to exclude instability. Parameters were chosen as follows: es = 0.99; /3 = 0.2; a = 0.1. We assume that the degree of apoptosis differs between stable and unstable cells. Stable cells have intact apoptosis (a = 0.5), while unstable cells have impaired apoptosis (a = 0.2). For (a) em = 0.1. For (b) em = 0.4. Low DNA hit rate corresponds to u = 0.07, and high DNA hit rate corresponds to u = 0.8. Fitness landscapes for successive mutants are given in Figure 7.4.

treatment more difficult in this scenario, (ii) The cost of generating lethal mutants in unstable cells must be higher than the cost of cell cycle arrest in stable cells. If this is not the case, it does not pay to retain repair mechanisms, and the fitness of unstable cells can never be reversed. In this case, treatment has a higher negative impact on stable than on unstable cells, and the mutators are resistant.

Chapter 8

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