Under optimum conditions, Escherichia coli doubles in cell number every 20 minut
ID: 200742 • Letter: U
Question
Under optimum conditions, Escherichia coli doubles in cell number every 20 minutes. Optimum conditions include the "right" temperature, food, pH, amount of oxygen, and amount of light. In the data below, samples were taken from a growing E. coli culture every 15 minutes for a total of 7 hours. The samples were diluted and grown on LB agar plates. Dilution factor CFU/mL Time (minutes) CFU 0 15 30 45 60 90 120 150 210 270 330 390 Plate 1 200 195 192 270 42 80 168 42 50 54 51 260 176 Plate 2 213 200 214 266 37 83 185 37 53 49 249 183 1) Determine the number of CFU/ml of the original culture for each time set. 2) Use the completed data to plot a growth curve of log CFU/ml vs time. Include an appropriate title and axis label. 3) Calculate the generation time and growth rate constant of the culture. 4) Does the term growth convey the same meaning when applied to bacteria and to multicellular organisms? (i.e: what is the difference between growth of bacterial culture vs multicellular organisms) Explain.Explanation / Answer
1. CFU/ml is no. of bacteria counted/dilution factor for examplefor 0 hours for plate set 1 it is 206.5/10-4 = 206,5000
In this case take the average of two CFUs,
2. After you have calculated CFU, calculate the log of CFU and plot is against time. log of CFU is on y axis whereas time is on x axis.
3. generation time = t(time per generation)/n (number of generations)
= 15/28 = 0.535
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.