IV. (4 points; each 1 point) Virtual growth curve a. Fill in the \"Viable Count\
ID: 257213 • Letter: I
Question
IV. (4 points; each 1 point) Virtual growth curve a. Fill in the "Viable Count" column of Table 2 below. b. Plot the absorbance versus time. c. Plot the log of the viable cells versus time on semi log paper Table 2. Number of Colonies of E. coli per time point Time in Number of colonies on plates 0.10 0.10 0.14 0.15 0.19 38 15 30 45 60 ND ND ND ND ND 90 ND ND ND T o 90 0.238 TNTC 280 ND 18 o chuurnt 120 0460 TNTC 38638 150 |0.701 |TNTC TNTC 91 0,920 TNTC 1.238 TNTC Tg.many 210 1 bcun 240 1.266 TNTCTNTC |TNTC |255 262 Note: To determine CFU/ml, 0.1 ml was plated d. Calculate the length of the lag period and the doubling time.Explanation / Answer
a. the viable count of bacteria implies the number of bacteria /ml of the diluted sample. We are given the various dilution factors and the respective absorbance of the particular sample. We can use the formula:
CFU/(volume of sample plated) X (dilution factor) = number of bacterial/ml
Volume of sample plated given = 0.1ml
Number of colonies/ml = 38/ 0.1 X 10-5 = 380 x 105
Number of colonies/ml = 0
Number of colonies/ml = 300/0.1 X 10-5 = 3 x 108
But the answer should be more than 300 million bacterial/ml
Note:Calculate all the viable cells under different dilutions and the final viable count will be the average of the three. In places where there are 2 values, take the average of the two.
Based on the observation: between 120mins to 180 mins there is a linear trend. Considering the two values:
OD1 = .460
OD2 = .701
T1 = 120 min
T2 = 150 min
Use the interpolation method
r = (ln [OD2-OD1)] / (T2 – T1)
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