ENUMERATION OF BACTERIA: 2314 Manual - Ex 19, p.55, 3054-Ex 21, p. 93 Wewill use
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ENUMERATION OF BACTERIA: 2314 Manual - Ex 19, p.55, 3054-Ex 21, p. 93 Wewill usethe yeast accharo ny escereis einsteadofbacteria.- We will also calculate the percent viability of the yeast suspension. Small Square First period .Follow the dilution and plating procedures outlined in the lab manual To calculate percent viability, it is necessary to obtain an estimate of the number of yeast cells per ml in the original yeast suspension. The procedure is called a direct count, and a special counting chamber is used. The drawing below shows the region of the counting grid to use for the count. Count the number of yeast cells in six small squares, and calculate the average number of cells per square. Count contiguous squares in rows to avoid bias in selecting squares. . Conversion Factor: Each small square accurately represents a volume of 1/(4x106) mls liquid in the counting chamber. Therefore, to calculate the number of cells per ml, use the following equation: No. cells per ml suspension Avg. no. cells per square x (4 x 10) where, 4 x 106 conversion factor for this counting chamber. 7 Second Period counter if necessary for plates with large numbers of colonies. Assume each colony arose from a single cell in the original suspension. . Count the yeast colonies on a plate that has between 30 and 300 colonies using the Quebec colony Calculate the number of viable cells per ml in the original suspension and percent viability of the cell suspension according to the following equations: No. Colony Forming UnitsCF.Ux Dilution Factor (see below) Units No. viable yeast cellsNo. Dilution plated No. viable yeast cells per ml. No. yeast celis per ml. (Direct count) x 100% a.Explanation / Answer
1) A) 0.4 X 104 = 4 X 103
B) 0.9 X 10-2 = 9 X 10-3
C) 18 X 10-4 = 1.8 X 10-5
D) 13.44 X 108 = 1.34 X 109
2) A) 3 : 30 = 1: 10 = 101
B) 1 : 99 = 1: 100 = 102
C) 0.5 : 49.5 = 1: 100 = 102
D) 0.1 : 99.9 = 1: 1000 = 103
E) 0.5 : 4.5 = 1: 10 = 101
3) 3 mls bacterial suspension plus 6 mls water gives a 1 : 3 dilution.
4) 4 ml of bacterial suspension is needed to make 1 : 9 dilution.
5) 1.8 mls bacterial suspension would be added to 9 mls broth to obtain 1 : 6 dilution.
6) a) The dilution in tube B is 101
b) The dilution in tube C is 102
7) a) Bottle C 103
b) Bottle C 104
8) 350 X 100 X 10 X 10 = 3.5 X 106
3.5 X 106 / ml viable organisms are present in stock, if dilution C was counted.
34 X 10 X 100 X 10 X 10 = 3.4 X 106
3.4 X 106 / ml viable organisms are present in stock, if dilution D was counted.
Stock solution contain 4.4 X 106
Direct count in dilution D = 4.4 X 106 / 105 = 44
Viable count in dilution D = 34
Percentage viability = 34 / 44 X 100 = 77.27 %
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