Run [H]t=0 [N]t=0 EtOH [H]t=end of run 100×[G]t=end of run/[G]t=0 100×[F]t=end of run/[F]t=0   (mmol/L) (CFU/L)   (mmol/L)     1C 833.3 G + 833.3 F 3.20*1011 no F = 52.2 -- 6.3% 2C 1111.1 G 1.25*1011 no -- -- -- 3C 1111.1 F 1.10*1011 no -- -- -- 4C 1666.6 G 8.92*1010 no G = 24.89 1.5% -- 5C 1666.6 F 1.38*1011 no F = 363.30   19.47% 6C 1111.1 G 5.75*1010 yes G = 188.0 17.2% -- 7C 1111.1 F 5.80*1010 yes F = 277.7 -- 25.13% 8C 555.5 G + 555.5 F 3.10*1011 yes G = 10.3; F = 176.4 1.9% 35.0% 1B 833.3 G + 833.3 F 8.30*1011 no G = 21.5; F = 110.10 2.6% 13.3% 2B 1111.1 G 1.68*1011 no -- -- -- 3B 1111.1 F 9.40*1010 no -- -- -- 4B 1666.6 G 8.89*1010 no G = 60.56 3.6% -- 5B 1666.6 F 1.10*1011 no F = 73.28 -- 4.4% 6B 1111.1 G 9.70*1010 yes G = 194.3 18.6% -- 7B 1111.1 F 5.10*1010 yes F = 289.06 -- 27.5% 8B 555.5 G + 555.5 F 5.2*1010 yes G = 34.67; F = 170.94 6.6% 34.9%

Table 1: Values of initial concentration of hexoses (H)t=0, yeasts (Nt=0), final concentrations of hexoses (H)t=endofrin and percentages of residual sugar for the experimental runs performed.