1. A grad student has a culture of E. coli with a total volume of 1,800 mL, and
ID: 52383 • Letter: 1
Question
1. A grad student has a culture of E. coli with a total volume of 1,800 mL, and CD reading at 550 nm of 1.7. He transfers 1 mL culture into 9 mL tryptone broth. Next, he transfers 0.1 mL of this dilution into 0.9 mL tryptone broth, and repeats this two more times to make a total of 4 serial dilutions. He then pipets 200 uL from the fourth dilution on an agar plate, spreads it over the surface, and puts the plate in an incubator to grow overnight. The next day, he counts 386 cfus (colony forming units) on the plate. a. What is the dilution in the third tube? Answer with the exponent to the power of 10, that is ? in 10? b. What is the dilution of cells spread onto the agar plate? Answer in decimal notation.Explanation / Answer
Answer
1. If the student transfers 1mL sample culture into 9 mL tryptone broth here first dilution (1/10) took place in the first test tube, then he transfers 0.1 mL of this dilution into 0.9 mL tryptone broth and second dilution (1/100) happens. Then, from this he draws some more culture to the third tube and here third dilution would take place (1/1000).
Then, the dilution factor in the third tube will be 10-3.
2. From the fourth dilution 200uL of culture is taken. Then dilution of cells spread of 200uL on the agar plates is contains 0.02uL bacterial cells.
3. Viable cells can be calculated as
Total dilution factor = 10* 100 =10^3
Total colony forming units = 386
386 Colony forming units/mL divided by 10-3 = 386 x 103 = 3.86 x 106 CFU/ml
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