Assuming that an E. coli inoculum contained 1 X 10^8 CFU/ml, all of the mutation
ID: 83044 • Letter: A
Question
Assuming that an E. coli inoculum contained 1 X 10^8 CFU/ml, all of the mutations that led to streptomycin resistance were in the E. coli culture at the time of inoculation, and any bacterial growth in any of the 1,000ug/ml streptomycin tubes came from one resistent bacterial cell per tube, what was the concentration of streptomycin resistant cells in the E. coli inoculum (in cells/ml)?
Tube 1: 0 ug/ml of streptomycin
Tube 2: 2ug/ml of streptomycin
Tube 3: 10ug/ml of streptomycin
Tube 4: 100 ug/ml of streptomycin
Tube 5: 500 ug/ml of streptomycin
Tube 6: 1000ug/ml of streptomycin
(Each tube contains 2ml LB broth and 5ul of E. coli culture.)
Explanation / Answer
Answer:
Number of E.coli CFU in inoculum = 1 x 108 CFU/ml
Tube content = 2ml LB Broth + 5ul E.coli culture +1000ug/ml of streptomycin
1ml (1000ul) of E.coli culture contains = 108 CFU
5ul of E.coli culture contains = (108*5)/1000 = 5 x 105 CFU / 2ml of culture = 2.5 x 105 CFU/ml
Any growth in this tube comes from single streptomycin resistant mutant cell. That means, streptomycin mutant frequency is 1 mutant / 2.5 x 105 CFU/ml
Number of mutants in original inoculum 1 x 108 CFU/ml = (1 x 108) / (2.5 x 105) = 1000/2.5 = 400
Therefore there would be ~400 streptomycin mutants in the E.coli inoculum.