If, after removal of the lowest dilution with all positive tubes, no dilution with all positive reactions remains, then select the lowest two dilutions and assign the sum of any remaining dilutions to the third dilution. In example E, the highest dilution with all positive tubes contains 10 ml; this dilution was removed in the second step. Four dilutions, none of which have all positive tubes, remain. Under these circumstances, select the two lowest remaining dilutions corresponding to 1 and 0.1 ml. sample for the third dilution, add the number of positive tubes in all higher dilutions (0.01 and 0.001 ml sample),to yield a final combination of 4-4-1.