ABSTRACT
Objective To investigate the relationship between the polymorphism of cytochrome P450 19 (CYP19) gene rs10046 and the risk of colon and rectal cancer.Methods Using polymerase chain reaction-restriction fragment length polymorphism (PCR-RFLP) to analyze gene polymorphism in CYP 19 gene rs10046 in 198 cases of colon and rectal cancer patients (case group) and 309 cases of healthy controls (control group).The genotype frequency and relative risk of CYP19 gene rs10046 between the two groups were compared and the relationship with the clinicopathological features of colorectal cancer was analyzed.Results In case group,the prevalence rates of CYP19 rs10046 genotypes C/C,C/T and T/T were 28.3％,44.4％ and 27.3％,respectively,and 17.2％,51.8％ and 31.1％ in the control group,respectively,with statistical significant difference (P ＜ 0.05).Compared with wild-type C/C,the susceptibility of colorectal cancer with the genotypes of C/T and T/T was decreased by 0.521 (95％ CI:0.330-0.822)and 0.532 (95 ％ CI:0.322-0.880) respectively.Moveover,in the non-smoking group,the risk of colorectal cancer with genotype T/T or C/T was decreased by 0.409 (95％ CI:0.210-0.798) compared with genotype C/C.The interaction was not exist in smoking group.Conclusions The polymorphism of CYP19 gene rs10046 is related to the susceptibility of colon and rectal cancer.The T/T,C/T genotype of CYP19 rs10046 decrease the risk of colon and rectal cancer,and which might be the protective factor of colon and rectal cancer.