In: Biology
Exercise 6: Determination of growth of E.coli
At the conclusion of this exercise you should be able to:
Description of the experiments
This experiment will determine the growth of E.coli over a four hour time period. In order to record this growth, we will be using two methods of determining growth: turbidometry and viable count methods. To begin this experiment, a 1.0 ml sample of an overnight culture of E.coli was placed in a 100 ml flask containing sterile nutrient broth. To determine an initial timepoint (time zero), a 2.0 ml sample was removed, 1.0 ml of which was used to examine turbidometry (Absorbance) at 600nm wavelength and 0.1 ml of the sample was applied to a nutrient agar plate and spread onto the plate to determine viable counts.
The flask was subsequently placed in a 370 C shaking water bath to allow the cells to grow. The flask was removed to obtain an hourly sample of culture, where 2.0 ml was aseptically removed from the flask to perform turbidometric and viable plate count methods (times 1 to 4 hours). To perform the viable counts over time, serial ten-fold dilutions of sample were made by adding 0.1 ml of sample to 9.9 ml of sterile water (this dilution is know as 1/10, or in scientific terms a 101 dilution). For this experiment, dilutions were made up to 107 for each subsequent timepoint.
A 0.1ml sample was removed from dilutions and added to a nutrient agar plate. The sample was subsequently spread onto the surface of the plate covering all quadrants of the plate (this is known as a lawn inoculum). After allowing plates to dry, plates were inverted upside down and placed in an incubator at 370C and removed after 18-24 hours. After incubation, plates showing colony numbers of 300 or less were counted for each time point. These are known as colony forming units (CFUs).
For a visual demonstration of how to perform serial dilutions and observe the appearance of colonies formed using this technique, watch How to Quantify Bacterial Culture Concentrations Using Serial Dilution and Plate Counts:
Results
Turbidometric recordings
Time (hours) |
Optical Density (Absorbance) OD600 (nm) |
0 |
0.005 |
1 |
0.03 |
2 |
0.21 |
3 |
0.29 |
4 |
0.35 |
Viable count (CFU) after 18 hours
Time (h) |
# Colonies on plate |
Dilution used |
# cells/ml * |
Log #cells/ml |
0 |
10 |
10-3 |
||
1 |
184 |
10-2 |
||
2 |
179 |
10-5 |
||
3 |
56 |
10-6 |
||
4 |
64 |
10-5 |
*Use proper scientific notation
# colonies on plate X 1/dilution X PF = #cells/ml e.g. 33 cells on the 10-5 plate would be:
33 X 105 X 10 =3.30 X 107 cells/ml (33 000 000 cells/ml)
Generation time =∆ t log 2
Log n – log N
Where: N = number of bacteria at a particular time point during log phase
n = number of bacteria at a second time point during log phase
∆t = time
ANSWER 1 :-
TIME (in hours) | Number of colonies on plate | Dilution used | Number of cells/ml | Log(Number of cells/ml) |
0 | 10 | 10-3 | 1 x 105 | 11.51 |
1 | 184 | 10-2 | 1.84 x 105 | 12.12 |
2 | 179 | 10-5 | 1790 x 105 | 19 |
3 | 56 | 10-6 | 5600 x 105 | 20.14 |
4 | 64 | 10-5 | 640 x 105 | 17.97 |
Cells/ml at 0 hour = (10) x 103 x 10 = 1 x 105 cells/ml
Cells/ml at 1 hour = (184) x 102 x 10 = 1.84 x 105 cells/ml
Cells/ml at 2 hour = (179) x 105 x 10 = 1790 x 105 cells/ml
Cells/ml at 3 hour = (56) x 106 x 10 = 5600 x 105 cells/ml
Cells/ml at 4 hour = (64) x 105 x 10 = 640 x 105 cells/ml
ANSWER 2 :-
NOTE :- Respected Sir/Madam, please specify the formula for generation time so that the last question can be answered as according to me, both the generation number and the change in the microbial population over time is important to be considered. Please provide an upvote if the answer seems satisfactory.