1 ml 1 ml 1 ml 1 ml 9 ml 9 ml 99 ml 9 ml 1 frm each tube, A, B, C, and D, was pl
ID: 148973 • Letter: 1
Question
1 ml 1 ml 1 ml 1 ml 9 ml 9 ml 99 ml 9 ml 1 frm each tube, A, B, C, and D, was plated onto each of 4 petri dishes. The number of colonies (CFUs) counted on each plate are as follows: A. TNTC (Too Numerous to Count) B. TNTC C. 350 D. 34 How many viable organisms per ml. were in the stock solution if dilution C was counted? If dilution D was counted? direct count (using a counting chamber) showed that the stock solution contained 4.4 x 10% r ml. Calculate the percent viability using the direct count and the viable count from bacteria per dilution D.Explanation / Answer
therefore total viable bacteria in C is 350 x 100 = 35000 CFU
it means 1 ml of B contains 35000 CFUs
therefore 10 ml of B contains 35000 x 10 = 350000 CFUs
it means 1 ml of A contains 350000 CFUs,
therefore 10 ml of A contains 350000 x 10 = 3500000 = 3.5 x 106 CFUs
It means 1 ml of stock solution contains 3.5 x 106 CFUs if dilution C was counted.
1 ml of D contains 34 CFUs,
therefore total viable bacteria in D is 34 x 10 = 340 CFU
it means 1 ml of C contains 340 CFUs
therefore 100 ml of C contains 340 x 100 = 34000 CFUs
it means 1 ml of B contains 34000 CFUs,
therefore 10 ml of B contains 34000 x 10 = 340000 = 3.4 x 105 CFUs
it means 1 ml of A contains 340000 CFUs,
therefore 10 ml of A contains 340000 x 10 = 3400000 = 3.4 x 106 CFUs
It means 1 ml of stock solution contains 3.4 x 106 CFUs if dilution D was counted.
The bacterial count should be 44 in 1 ml of D, however, it is given that 1 ml of D contains 34 CFUs.
It means D has viable count of 34 CFUs out of 44 total counts.
So to calculate the percentage viability, (34/44) x100 = 77.27%
Therefore the percent viability will be 77.27%
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