Eating 20 to 40 grams per day of broccoli, cauliflower, and Brussels sprouts could reduce the risk of colon cancer by 20%, according to a new study published in *BMC Gastroenterology*. Colon cancer is the third most common cancer and the second leading cause of cancer-related deaths worldwide, with approximately 1.9 million new cases diagnosed annually. Diet is among the most influential factors affecting the risk of developing this cancer.
The research, conducted by epidemiologists from the University of Mongolia, is a recent systematic review of multiple published studies, encompassing a total of 639,539 participants and 97,595 cases of colon cancer. The analysis revealed that consuming 20-40 grams of cruciferous vegetables daily offers the optimal protective effect per gram.
In addition to being a rich source of beneficial phytochemicals—including flavonoids, fiber, vitamin C, and carotenoids—cruciferous vegetables are also abundant in glucosinolates, compounds known for their anti-cancer properties. When chewed, these compounds break down into bioactive isothiocyanates, most notably sulforaphane (SFN). This molecule is responsible not only for the vegetables’ strong and characteristic odor but also for their cancer-preventive effects.