(9) 2505-5 fraction 5 [pUte 2] 50% drqiUcrment u 20 or 025% of sample at this Rage (20 x Oi)I259b). 



Therefore, 025%  895 pgram equivlaents in fraction 5 from the TLC plate by oompasion with standard 

 curves. 



100%yD25% - 400. Tielding (895 x 400) - 358.000 pg. or 358 ng in total sample. 



That 358 ng is equivalent to 91.8% of the total sample (100% x 0.958 x 0.958 * 91 J%). 



Therefore, the sample was calculated to contain 390 ng birvetoxin (358 x 100%^1.8%). 



* Based on a weight of 38i)36 g bver supplied, appimlm ately 102 ng toxin per g bver.* 



132 



