If serial dilutions were performed, and the number of colonies were expected to decrease by a factor of 10 with each higher dilution, what could cause the results to be other than those expected?

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Reid WolffLv2

19 Mar 2018

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Viable cell counts are often performed on exponentially growing cells in order to relate the absorbance reading (AU) with actual cell number. For this purpose, a sample of the culture is removed during the exponential phase after the absorbance reading is recorded. Spread plates are prepared in order to perform a viable cell count. The goal is to have a spread plate containing between 30 and 300 cells (colonies) when plating a volume of 0.1 ml (100 Î¼l). To determine the correct dilution, you will need to estimate the number of cells/ml present from the AU (absorbance units) value of your sample. For this calculation, we will assume that 1 AU is equivalent to 10⁷ cells /ml. So:

AU of sample x [(10⁷ cells/ml)/1 AU] x 0.1 ml plated = total # of cells expected

You want to calculate two different dilutions that will give you between 30 and 300 colonies (cells). In order to determine the maximum dilution factor, divide the total # of expected cells by 30. In order to determine the minimum dilution factor, divide the total # of expected cells by 300. For example if the AU was 0.3, the total number of expected cells would be 3.0 x 10⁵. In this case the maximum dilution factor would be 10⁴ (3.0 x 10⁵/30) or a dilution of 10^-4 and should produce approximately 30 colonies. The minimum dilution factor would be 10³ (3.0 x 10⁵/300) or a dilution of 10^-3, which should produce ~300 colonies. Since performing a single dilution of 10^-4 or 10^-3 would be highly inaccurate, serial dilutions are performed. Typically, the spread plates are prepared in duplicate for greater accuracy.

QUESTION 1:

You have removed a volume of a culture of E. coli during exponential growth in order to measure the viable cell count. If the absorbance reading was 0.3, estimate the maximum dilution factor you would need in order to obtain between 30 and 300 colonies on a plate spread with 0.1 ml of the dilution. Assume that there are 10⁷ cells/ml in one absorbance unit.

QUESTION 2:

You have removed a volume of a culture of E. coli during exponential growth in order to measure the viable cell count. If the absorbance reading was 0.3, estimate the minimum dilution factor you would need in order to obtain between 30 and 300 colonies on a plate spread with 0.1 ml of the dilution. Assume that there are 10⁷ cells/ml in one absorbance unit. â

bluealligator624

You perform a series of serial dilutions. You decide to plateeach of your dilutions using the spot plate method because your labis broke and you are trying to conserve plates. You set up 8 serial10-fold dilutions from an unknown sample. After plating three 25 Âµlaliquots of each dilution and allowing for proper incubation, youcount 10, 15, and 16 colonies in your 10^-6 quadrant, and5, 7, and 4 colonies in your 10^-7 quadrant. In the10^-3 quadrant, the growth is very dense, but you thinkyou count 42, 34, and 41 colonies. (a) What is the cfu/ml of thisculture? (b) Which plate did you choose for your calculation andwhy?

apricotmallard650

You are trying to determine the concentration of a bacterialculture. You set up 7 1/10 serial dilutions of you culture byadding 1 ml of culture from the more concentrate sample into 9 mlof broth. You then plate 0.1 mL of each dilution on an agar plate.The next day you observed growth on each plate. The plate that camefrom the 7th dilution tube contain 12 colonies. The plate that camefrom the 6th dilution contained 135 colonies. The plate that camefrom the 5th dilution contain 1200 colonies (that was a lot ofcounting). All other plates contain lawns of bacteria (to many tocount: TMTC). Report the concentration of the original culture.