SHEET Lab Section 6-1 I was present and performed this exercise (initials) (cont
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SHEET Lab Section 6-1 I was present and performed this exercise (initials) (continued) 23 You are told that a sample has between 2.5 x 103 and 2.5 x 10s cells/ml. Devise a complete but efficient (that is, no extra plates!) dilution scheme that will ensure getting a countable plate. 24 A sample has between 3.3 x 104 and 3.3 x 106 CFUhmL. Devise a complete but efficient (that is, no extra plates!) dilution scheme that will ensure getting a countable plate. 25 Two plates received 100HL from the same dilution tube. The first plate had 293 colonies, whereas the second had 158 colonies. Suggest reasonable sources of error. Two parallel dilution series were made from the same original sample. The plates with sample volumes of 26 10 5 ml from each dilution series yielded 144 and 93 colonies. Suggest reasonable sources of error. SECTION 6 Quantitative Techniques 477Explanation / Answer
Ans. 24. An optimum CFU count is preferred to be 100 -300 CFU count per plate under most conditions. A CFU count lower than 100 can be accounted but those above 300 are not preferred because the colonies are too closely placed.
So, the upper limit of CFU count = 300 CFU.
#1. To get a 300 CFU count, the volume of inoculum is usually taken to be 100 uL (= 0.100 mL).
So, the CFU count of ‘ready-to-inoculate’ dilution =
300 CFU / 0.1 mL = 3000 CFU/ mL
That is the concentration of final diluted solution is 3000 CFU/ mL, you get a 300 CFU count for the plate.
#2. Required dilution for getting 3000 CFU/ mL =
(300 CFU/ mL) / (3.3 x 106 CFU) = 9.091 / 104
That is, dilute 9.091 mL of the stock culture to 104 mL
#3. Serial dilution
First dilution: Take 1.0 mL of stock culture and mixed it with 9.0 mL sterile diluent (distilled water of buffer or saline). The final volume is 10.0 mL.
Dilution factor = Volume of stock culture / Volume of final solution
= 1.0 mL/ 10.0 mL
= 1 : 10 = 1:101
Second, and Third: Make 1:10 serial dilution similarly to get a final dilution of 1: 103 in tube three or third dilution.
So, in tube three, the dilution = 1: 103
Final dilution: We have 1: 103 dilution in tube 3.
We have to get 9.091 / 104 dilution that serves our purpose.
Take 9.091 mL of third dilution and add to 0.909 mL sterile diluent in tube four. The current/ immediate dilution = 9.091mL / 10.0 mL = 9.091: 10.0
Total dilution in tube 4 = Current dilution x dilution till tube 3
= (9.091: 10.0) x (1: 103)
= 9.091/ 104
So, the final dilution of tube 4 = 9.091/ 104
#4. Inoculation
We have, the final dilution of tube 4 = 9.091/ 104
Concertation of Tube 4 = 3000 CFU/ mL
Inoculate 100 uL of tube 4. It shall give a CFU count of 300.
A CFU count below 300 is countable as well spread colonies even if the concertation is at the lowest extreme.
Ans. 26. The possible sources of error are-
I. Pipetting error- Pipetting a small amount of stock solution for preparing serial dilution may be a possible reason. Pipetting less or more than the actual quantity at any point of serial dilution and inoculation gives lower or higher CFU count than the actual value.
II. Non-homogenous mixing- Non-homogenous mixing of stock or previous or current dilution in a serial dilution may cause un-even distribution of microbes in the final solution- many microbes may be aggregated and some cells will be present as single cell; or some portion of the solution has few cells while the other has large number of cells. One colony is considered a CFU. When cells are present in aggregates or NOT present as single cells in the solution, many cells clumped together form one CFU. Thus, aggregates/ clumps give less CFU than if all of its cells were present as separate single cell. A random inoculum consisting of larger concertation of cells than rest part of a non-homogenous solution may give higher CFU count.
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