Question

Now knowing the MPN value, you decide to do a dilution series in order to figure out the original bacterial concentration of the Swimming Hole, just in case it is not a coliform causing the gastric distress and simply a bacterial overgrowth. You perform the following dilution set: from your 16.3-liter Swimming Hole sample you transfer 10.7 x 10-3 liters into container A which has a volume of 6,250ul, you then transfer 1.7 mL from container A into container B which has a volume of 4.11 x 103 yl of water. The third transfer is 265 yL from container B into container C which contains 2336 l of water. The fourth transfer is 1.1 x 10-3 L from container C into container D which contains 540 y of water. Lastly in order to test your nonstandard dilution you plate 233 L from each container (A, B, C, D) onto a Nutrient Agar. The schematic below indicates the growth you observe after 48 hours of incubation. 16.3 Plate 1 Plate 2 Plate 3 Plate 4 Plate 5 After 48 hours of Incubation at 37° C you observe the following growth on plate 5: Page 12 of 15 Plate 1 Plate 2 Plate 3 Plate 4 Plate 5 345,000 colonies JL 117,000 colonies 55,000 colonies JL 5,000 colonies L 270 colonies 71. What was the original microbial concentration of the Swimming Hole sampled, and how much bacteria was present in the initial sample volume obtained? (5 pts, show your work for credit) 72. Just to confirm your suspicions that the gastric disease may be coming from the Swimming Hole, and that the water may have a high coliformis count responsible for the gastric ailment, you move on to the confirmation test of the 3-test process that is the typical EPA Bacteriological Examination of Water testing process and inoculate an EMB plate. You find that after 24 hours at 37-C it appears to have growth indicative of Escherichia coli. Describe this growth as it will become an important parameter to include in your report due back to the CDC. (3 pts) 73. Considering that your community is concerned about the possible coliformis epidemic, which value would you report at a town-hall meeting (the MPN or the non-standard SPC dilution result) and why? (2 pts)

Answer 1

so final concentration of stock is 12852 colonies /ml .

to t4153L 67 1 065 m 116 16.3 233 me ) 233 7/233 eu 8/203 11x1834 265 ne Volume 107me 1.7 mi 1100 m. Volume – 107x10² 17com 265 m Total (1100+540) 6250 +187680) (1700) une - 410 au (265) 2380 ure loro 5310 2601 17000 265 - factor 6113,250 107000 = US 3.41 9.81, 1.49 Total diwtion a 1.05 x 3.49 x 9.81 x 1.49 52.36. so we of 52.36 fold doluded Solution contain 270 Colonies. 1600 eur - 295 Rio 1000 . . ford will = 246 Gloning have = 2460 52.36 = 12852 Glonies 107000 ne of - 12852 & 10t 800 = 1.378 109 Colonies,

And 1.37*10^9 colonies are present in initial volume obtained which was 10.7*10^-3 Litre.

Thank you