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BIOL240 Lecture Notes - Lecture 14: Serial Dilution, Growth Medium

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30 Oct 2017
School
University of Waterloo
Department
Biology
Course
BIOL240
Professor
Cheryl Duxbury

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Document Summary

Direct count -> total number of bacteria (viable + dead) Quantitative plating -> determine population count of a culture of organisms. Standard plate count (spc) -> counts viable organisms; colonies that are seen on plates after incubation (used to determine # of organisms in water, milk, and food) Typical plate-count procedure -> combination of a serial dilution and use of suitable growth medium to detect colonies from aliquots of the dilutions, each viable cell will develop into a colony. Colony forming units (cfu) instead of colonies in a quantitative plate count (number of. How to calculate cfu/ml is = number of colonies counted on plate by the dilution factor.

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Related Questions

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 μl). 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. ​

Determining Dilution Scheme




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 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 ?l). 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.



(NOTE: Do not put units in your answers, if you do the computer will grade the answer as wrong and you will lose the points. Answers should be written in numerical form or if you want to write as scientific notation you can write as 1.22E8 or 1.22E-8. The system will only take up to 15 significant figures.) (Another NOTE: Your answer must be within 5% of the correct value to get credit for this problem.)



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 0.8, 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. Answer:



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 0.8, 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. Answer:

FOR #4 A & B I don't know if I did it correctly.....Please check & the other questions I'm having a heard time on. Thanks!

A. You have been given a tube of E. coli. You are asked to make 1 mL total volume of 10-1 dilution of the bacterial culture. Explain how you would do this. Show all necessary calculations.

____ ml cells + _____ ml water = 1 ml (total volume)

V1 D1 X V2 D2
1 mL X 1 = V2 X 10^-1
V2 = 1 mL/10^-1 so V2 = 10 mL.
But because V1 is part of V2. The answer is: 10 mL-1 mL= 9 mL of diluent must be added.

V1 D1 X V2 D2
1 mL X 1 = V2 X 10^-2
V2 = 1 mL/10^-2 so V2 = 100 mL.
But because V1 is part of V2. The answer is: 100 mL-1 mL= 99 mL of diluent must be added.

You have bacteria at a concentration of 5 x 108 CFU/mL. You spread 1 mL of this sample on an agar plate to obtain isolated colonies. How many colonies do you expect to find the next day after incubation at 370C? Can you count these colonies? Can you use this plate for determining concentration?

A. You have bacteria at a concentration of 1 x 103 CFU/mL (in real life – you don’t know this, but we are just working on math skills here). You transfer 1 mL of this sample into 9 mL of water and then spread 1 mL on a plate of agar. How many colonies do you expect to find the next day after incubation at 370C? Can you count these colonies? Can you use this plate for determining concentration?

You take 0.05 mL of a culture of bacteria at a concentration of 4 x 107 CFU/mL, and add 4.95 mL of water to it. What is the dilution that you have performed? What is the concentration of bacteria (CFU/mL) in the diluted culture?

A. You have diluted a sample by 1000 fold (1/1000) and plated 1 mL on an agar plate. You observe 55 colonies. What was the concentration of the original sample in CFU/mL?

A. A bacterial sample has a concentration of 3 x107CFU/mL. You make serial dilutions of 10-3 followed by 10-2 and 10-1 dilutions. You finally plate 1 mL of the last dilution on an agar plate and incubate it at 370C. What is the total dilution? How many colonies do you expect to see on the plate?

Total dilution:

# of colonies expected on plate:

C. This time, you see 10 times fewer colonies than you had expected to see.What could have gone wrong? How will you fix this problem?

A bacterial sample has a concentration of 2 x106 CFU/mL. Show a scheme of dilutions to obtain 30-300 colonies on a plate. Your scheme should contain the volume of diluent (sterile water), volume of sample transferred each time, the concentration of bacteria (CFU/mL) in each dilution. You can assume you are plating 1 mL of the dilution on plates of nutrient agar. See Figure 1 in protocol.

These are the results of the experiment described in the protocol. From the data shown in the results section of Standard Plate Count protocol, calculate the concentration of bacteria (CFU/mL) in the original sample of E. coli. Show your calculations. Hint: remember that we don’t use plates that have less than 30 or more than 300 colonies to do these calculations.