A mixture of Escherichia coli and Halobacterium halobium was diluted in a 15% (w/v) NaCl solution...

 

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A mixture of Escherichia coli and Halobacterium halobium was diluted in a 15% (w/v) NaCl solution and plated on two types of growth media. E. coli will grow on medium with 0.5% NaCl but not on medium with 25% NaCl. In contrast, H. halobium will grow on medium with 25% NaCI but not on medium with 0.5% NaCl. After incubation at 30°C, the number of colonies on the plates were counted. The results obtained are shown in the following table. CFU H. halobium Volume plated # Colonies on # Colonies on CFU Total 0.5% NaCl 25% NaCl plates E. coli CFU/mL mixture plates per mL per mL 0.1 mL of 102 TNTC TNTC 0.1 mL of 10 TNTC TNTC 0.1 mL of 10* 280 70 a. Calculate the average number of CFU/mL of H. halobium and E. coli in the mixture. Based upon these calculations, fill in the blank boxes in the table above. b. A direct visual count of the original mixture indicated that contained 2 x 100 H. halobium cells/mL and 10 x 10 E. coli cells/mL. Do these results agree with the cell numbers calculated above from the plate counts? If so suggest a plausible explanation. (Assume that all manipulations were carried out properly and that E. coli cells can survive exposure to the diluent.) c. What do you predict would happen if the original dilution had been made in standard saline (0.85% NaCl)? A mixture of Escherichia coli and Halobacterium halobium was diluted in a 15% (w/v) NaCl solution and plated on two types of growth media. E. coli will grow on medium with 0.5% NaCl but not on medium with 25% NaCl. In contrast, H. halobium will grow on medium with 25% NaCI but not on medium with 0.5% NaCl. After incubation at 30°C, the number of colonies on the plates were counted. The results obtained are shown in the following table. CFU H. halobium Volume plated # Colonies on # Colonies on CFU Total 0.5% NaCl 25% NaCl plates E. coli CFU/mL mixture plates per mL per mL 0.1 mL of 102 TNTC TNTC 0.1 mL of 10 TNTC TNTC 0.1 mL of 10* 280 70 a. Calculate the average number of CFU/mL of H. halobium and E. coli in the mixture. Based upon these calculations, fill in the blank boxes in the table above. b. A direct visual count of the original mixture indicated that contained 2 x 100 H. halobium cells/mL and 10 x 10 E. coli cells/mL. Do these results agree with the cell numbers calculated above from the plate counts? If so suggest a plausible explanation. (Assume that all manipulations were carried out properly and that E. coli cells can survive exposure to the diluent.) c. What do you predict would happen if the original dilution had been made in standard saline (0.85% NaCl)?

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a b c of colonies on 05 NaCI plate Volume of colonies CFU E CFU H Total ... View the full answer