High calcium consumption may confer a reduced risk of colorectal cancer.1,2 Dai and colleagues3 recently reported in a case-control study that intake of calcium may be associated with a decreased risk of colorectal adenoma only when the dietary calcium:magnesium intake ratio is low. This finding provides one possible interpretation for
inconsistencies in previous studies of the association of calcium intake with risk of
Belonging to the same family in the periodic table, calcium (Ca2+) and magnesium (Mg2+) share the same homeostatic control system and have the potential to antagonize each other physiologically.5 A high calcium intake reduces absorption of both magnesium and calcium,6 whereas moderate magnesium deprivation results in negative magnesium balance but increased calcium retention7. Due to the potential competition between magnesium and
calcium, we hypothesized that the dietary calcium:magnesium ratio may modify the effects of calcium supplementation on colorectal carcinogenesis.
We sought to test this hypothesis using data from a randomized clinical trial of calcium supplementation (1200 mg/day) to prevent adenoma recurrence over a 4-year period. The outcome measure in this analysis was the recurrence of adenomas during the pre-specified main risk period (i.e. after a year-1 colonoscopy up to and including a year-4 examination)2. A validated semi-quantitative food frequency questionnaire (FFQ) was given at study entry
to assess usual diet.
Consistent with our hypothesis,3 we found suggestions that the baseline dietary calcium:magnesium intake ratio modified the effect of calcium treatment on adenoma recurrence. Among subjects with the intake ratio above the median, calcium supplementation had no effect on the risk of one or more recurrent adenoma (relative risk [RR]=0.98 [95% confidence interval (CI)=0.75–1.28]) (Table). In contrast, among those with the baseline ratio less than or equal to the median, calcium treatment was associated with reduced risk (RR=0.68 [95%CI=0.52–0.90]; test for interaction, P= 0.075).
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