T-Mobile Wi-Fi11:06 AM a instructure-uploads.s3.amazonaws.co C 2 of 2 Qeestions
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T-Mobile Wi-Fi11:06 AM a instructure-uploads.s3.amazonaws.co C 2 of 2 Qeestions l. what is the dilution factor fr dilutace? (Added-ounttotal amount-altaon factor) 2. What is the TOTAL dlution factor for Plate A? What makes the total dilution factor foer Plate B diflerent from Plate A7(Total dilution factoe individual dlution factors muhiplied by each other) J. Afler the plates were incubated at 30'C for 24 hrs, 273 colonics Sormed on platk A and 30 colonies fomed on plate B. What is the concentration,in c.f.u/ml, of the original diluted dint sample? As shown, the student has 5 mLs of the original culture. How many e.f.u'g dint are there in the samgle? S. Say the stadent changed dilution 5; adding 1 mL of the previous culture to 4 mls of sterile PBS A. How does this changpe the dlution factor of dilution 57 B. How does this change the TOTAL dilution factor? C. What would be the new original dist sample coecentration, ncfuig dit, if Plate A and Plale B yielded the same amount of colonics as indicated in quetion 3 Please attach and show all werk te receise full eredit.Explanation / Answer
Tube 1 dilution factor = 10^-2 = 100 fold dilution
Tube 2 dilution factor = 10^-1 = 10 fold dilution
Tube 3 dilution factor = 10^-1 = 10 fold dilution
Tube 4 dilution factor = 10^-1 = 10 fold dilution
Tube 5 dilution factor = 10^-1 = 10 fold dilution
Total dilution factor = 10^-6 = 1000000 fold dilution
On plate A, 1 mL of from tube 5 is inoculated. But, one plate B, only 0.1 mL of from tube 5 is inoculated.
Number of colonies on plate A = 273
Concentration of original undiluted sample = 273 X 10^6 CFU/mL
Number of colonies on plate B = 30
Concentration of original undiluted sample = 30 X 10^6 CFU/0.1mL = 300 X 10^6 CFU/mL
Total volume of undiluted sample = 5 mL
Total CFU = [(273+300)/2] X 10^6 = 2.86 X 10^8 CFU
If 1 mL of sample from tube 4 is added to 4 mL of tube 5,
Individual dilution factor = 0.5 = 5 fold
Total dilution = 100000 X 5 = 500000 fold dilution
Number of colonies on plate A = 273
Concentration of original undiluted sample = 273 X 5 X 10^5 = 1.365 X 10^8 CFU/mL
Number of colonies on plate B = 30
Concentration of original undiluted sample = 30 X 5 X 10^5 = 1.5 X 10^7 CFU/mL
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