8. (3 points; each 1 points) You were given 50 ml of sewage and asked to determi
ID: 255387 • Letter: 8
Question
8. (3 points; each 1 points) You were given 50 ml of sewage and asked to determine the number From this mixture, you took 1 ml and put it into 99 ml of sterile water. From the previous mixture, you took 10 ml and put it into 10 ml of sterile water. From the previous mixture, you took 1 ml and put it into 9 ml of sterile water. From the previous mixture, you took 5 m and put it into 45 ml of sterile water. Then, you plated 0.2 ml of the final solution, 0.1 ml and 0.01 ml, a n triplicate. The plates inoculated with 0.2 ml and 0.1 ml of the solution had TNTC (too numerous to count) and the 0.0 ml plating generated 51,49 and 32 colonies a. What is the final fold-dilution for the diluted solution used for plating? b. What are the bacterial counts per ml of original sample? c. What are the total bacteria counts in the original sample?Explanation / Answer
Initially, 5 ml of sewage water was mixed with 45 ml of sterile water (i.e. 1:10 dilution)
Next, from above mixture 1 ml sample was mixed with 99 ml of sterile water (i.e. 1:100 dilution and resulting dilution became 1:1000)
Then, 10 ml of above mixture was mixed with 10 ml of sterile water (i.e. 1:2 dilution and resulting dilution became 1:2000)
From the above mixture 1 ml of solution was taken and mixed with 9 ml of sterile water (i.e. 1:10 dilution resulting dilution became 1:20000)
Finally 5 ml of the above mixture was mixed with 45 ml of sterile water (1:10 dilution and resulting dilution became 1:200000)
a. Therefore, 200000 fold dilution is used for the plating.
0.2 ml, 0.1 ml and 0.01 ml were plated. 0.2 ml and 0.1 ml of the solution had TNTC, so we couldn't calculate colonies from those plates. 0.01 ml plating gave 51, 49 and 32 colonies. On average (51+49+32)/3 = 44 counts. Therefore from 0.01 ml we got 44 numbers of colonies,; so bacterial counts per ml of diluted sample be (44 X 100) = 4400 counts.
b. Therefore bacterial count/ml of original sample be (4400 X 200000) = 880000000 = 8.8X108 & c. Total bacterial count be 50 X 8.8 X 108 = 440 X 108 = 4.4 X 1010
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