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I need help with this please. Thank you. 4. If given a mixed sample containing E

ID: 265700 • Letter: I

Question

I need help with this please. Thank you.

4. If given a mixed sample containing E.coli and M. luteus. Which of the following two m edia, EM B or MSA, would be the most useful in the isolation of the above gram negative organism a explain you answer. (10 pts) Calculate the number of bacteria per ml of undiluted culture by using the follow ing data: (5 pts) Plate 1; had 115 CFUs and received 1.0 ml of the 1:1,000,000 dilution (1.0 x 10%) Plate 2; had 175 CFUs and received 1.0 ml of the 1:10,000,000 dilution (1.0x 10) Plate 3, had 28 CFUs and received 1.0 ml of the 1:100,000 dilution (1.0 x 10) Plate 4; had 317 CFUs and received 1.0 ml of the 1:100,000 dilution (1.0x 10) cells/ml of undiluted culture. Complete the following problems Show your work. (10 pts) i. Dilute a culture containing 5.0 x 10 cells/ml such that a 0.1 ml plating will give 20 colonies.

Explanation / Answer

The E. coli is a gram -ve bacteria which belongs to gamma-proteobacteria class. The M.luteus is a gram +ve bacteria which belongs to family micrococcaceae. The gram -ve bacteria are best characterized by EMB agar or broth. The gram+ ve are best characterised by MSA. The MSA agar supports only those who ferment mannotol and form colonies on the plate. It also has high concentration of salt which inhibits gram -ve bacteria.
EMB agar is a selective media. it contians dyes eosin and methylene blue that are toxic to gram +ve bacteria.

Colonies per cell = no. of colonie * dilution factor / volume plated.
         plate 1       = 115 * 10^6 / 1
                           = 1.15 * 10^8 /ml
plate 2 : 175 * 10^7 /1 = 1.7* 10^9 /ml
plate 3:   28 * 10^5 /1 = 2.8 *10^6
plate 4: 317 * 10^4 /1 = 3.17 * 10^6

Average of all the plate readings = 4.66 * 10^8 / ml

1. Sample = 5 * 10^9 / ml : Volume plated = 0.1 ml;   colonies = 20; dilution = ?

Sample = colonies * dilution / volume
5 * 10^9 = 20 * d / 0.1
20* d = 5 * 10^8
d = 5/20 * 10^8
d = 0.25 * 10^8
d = 2.5 * 10^7
SO, atleast 10^7 times or (1:25000000 ) times the sample has to be diluted .