Colorectal cancer is genetically well defined, and is known to progress through a defined series of stages, but what is the initial event that sets cells along this tumorigenic pathway? Mutation of the APC tumour-suppressor gene is certainly an early event, and chromosomal instability (CIN) is also thought to be a driving force of tumorigenesis, but whether CIN occurs early in tumorigenesis — perhaps even causing loss of heterozygosity of the second APC allele — is, at present, unknown. Martin Nowak et al., reporting in Proceedings of the National Academy of Sciences, have now devised a stochastic mathematical model to investigate this question.

They started by considering the six states, with respect to APC inactivation and CIN, that any cell in the colonic crypt could be in. Cells could have none, one or two functional APC alleles, in the presence or absence of CIN. They then determined how, and with what rate, a cell could progress from one state to the next by considering parameters such as mutation rate, loss of heterozygosity rate in normal and CIN cells, reproductive rate of APC−/− and CIN cells, and the number of dominant CIN genes.

They calculated the probability that the system reaches the state X2 (APC inactivation occurs in the absence of CIN) or Y2 (APC inactivation occurs in the presence of CIN) first, and therefore whether CIN could cause inactivation of the second APC allele. For the second scenario to occur, the number of CIN genes must exceed a certain threshold value. This number depends on several factors, including the selective cost of CIN and the effective number of stem cells per crypt. For a wide range of realistic parameter values, the crucial number of CIN genes is as low as 1–10.

So, under certain conditions, and if the number of dominantly acting CIN genes in the human genome exceeds a certain number, it should be possible for CIN to inactivate the first tumour-suppressor gene in colorectal cancer. The more time-consuming process — confirming this hypothesis — remains to be undertaken.