Question

11. A package of yeast (7.4g) was suspended into 100 ml of water. The

suspension was serially diluted by transferring 0.1 ml aliquots successively to 9.9

ml of water three times in secession. 0.1 and 0.2 ml aliquots were spread on

glucose nutrient agar. Colonies appeared as follows: 0.1 plate: 179, 0.2 plate: 32

1. How many yeast were there originally in the package? How many yeast

Were in a gram of material?

Please explain in detail and show all work.

Answer 1

Dilution factor (DF) = Initial volume (V_i) / Final volume (V_f)

In problem we diluted 0.1 ml aliquots to 9.9 ml of water.

Therefore, DF = 0.1 / 10 = 1 / 100 (Here final volume = 9.9 + 0.1 = 10 mL)

The dilution factor after three dilutions is-

DF = (1 / 100) x (1 / 100) x (1 / 100) = 1/100,00,00

Original concentration = 7.4 g / 100 mL = 0.074 g / mL

The concentration after three dilutions = (0.074) x 1/100,00,00 = 7.4 x 10^-8 g/mL

Number of yeast in the original package = number of colonies growing on plate x (1/volume used to put bacteria on plate) x (1/dilution)

From the 0.1 mL plate we find the number = 179 x (1/0.1) x (1/ (1/100,00,00)) = 179 x 10⁷ cells

From the 0.2 mL plate we find the number = 321 x (1/0.2) x (1/ (1/100,00,00)) = 160.5 x 10⁷ cells

From the 0.1 mL plate we find the number of yeast/g of material = (179 x 10⁷) / 7.4 = 24.19 x 10⁷ cells/g material

From the 0.2 mL plate we find the number of yeast/g of material = (160.5 x 10⁷) / 7.4 = 21.69 x 10⁷ cells/g material