Incubation of [4-14C, 1a-3H] 25(OH)D3 (0.7 mg; 3H/14C = 5) with
the homogenate of 1.9 g liver from vitamin D-deficient rainbow
trout (body weight = 240 g) for 4 h at 17 converted 61% of the
substrate to a more polar product with a 3H/14C ratio of 2.56, thus
confirming by the loss of the 1a-tritium atom, its identity as 1,25
(OH)2D3 (Fig. 3).