1. Serial Dilution Calculations Dilution Plating Questions PDF

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Question to answer during pre-reading for Lab 2: Suppose you do a four-step serial dilution with 100-fold, 10-fold, 10-fold then 5-fold dilutions. What will the Total Dilution Factor (TDF) be? Circle the correct answer. [Hint: see the previous page]. A)

5 x 105-fold

B)

5 x 104-fold

C)

5 x 103-fold

D)

125-fold

0.1ml

Our dilution plating scheme 0.1ml 1.0ml

1.0ml

A

B

C

D

9.9ml

9.9ml

9.0ml

9.0ml

100 fold

10 fold

10 fold

102

104

105

106

0.1ml

0.1ml

0.1ml

0.1ml

Undiluted Yeast

Dilution Factor = V2/V1

100 fold

Total Dilution Factor = DF x DF ect

PRACTICE QUESTIONS ! !

Serial dilution Calculations You need to identify how many steps are involved. It is easiest to find the dilution factor for each individual step and multiply these together TDF = DF1 x DF2 x DF3 x ….. and then work out what is asked in the question.

Question 1 ! ! !

1. What concentration will result from four consecutive 5-fold dilutions of a stock solution of 138 mg/ml?

Steps i) Calculate the total dilution factor for this question, showing your working.

ANSWER: TDF = 625 FOLD

ii) Would the 138 mg/ml stock solution be C1 or C2?

ANSWER = C1

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Now solve the question using DF = C1/C2

0.22mg/ml

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QUESTION 2 What three serial dilutions would generate a total dilution factor of 400-fold? Describe how you would do each dilution. (Hint: There is more than one possible answer) Many possibilities: The dilution factors need to be any three numbers (not 1) that multiply together to make 400. Then work out how each DF can be made. For example: 10 x 10 x 4 = 400

Two 10-fold dilutions (such as 1 ml sample + 9 ml diluent) and a 4-fold dilution (such as 1 ml sample + 3 ml diluent) would achieve this.

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QUESTION 3 ! !

2 ml of a solution is added to 4 ml of diluent. Then, 2 ml of the diluted solution is added to 4 ml of diluent. This dilution procedure is repeated four more times (i.e. a total of six dilution steps), each time using the previously diluted solution. What is the total dilution factor?

729 fold (= 36). Each DF is 3 and there are 6 successive dilutions.

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QUESTION 4 4. Three 4-fold dilutions of a stock solution produces a concentration of 1.5 mg/ml? What is the concentration of the stock solution?

TDF = 64, so C1 = 96 mg/ml

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QUESTION 5 5. Three 6-fold dilutions of a stock solution produces a concentration of 2.3 mg/ml. What is the concentration of the stock solution?

TDF = 216 so C1 = 496.8 mg/ml

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B. Dilution Plating Calculations Dilution plating experiments Often it is useful to draw a visual representation of the dilution scheme, in order to understand all the consecutive steps and calculate the total dilution factor.! ! !

QUESTION 1 1. A dilution plating experiment was carried out as follows: Step 1 - 0.1 ml of undiluted culture was mixed with 9.9 ml diluent. Step 2 - 0.1 ml of the diluted culture was mixed with 9.9 ml diluent. Step 3 - 0.1 ml of the serially-diluted culture was mixed with 2.9 ml diluent. Step 4 - 0.2 ml of the final dilution was spread on a nutrient agar plate. This plate was incubated at 37˚C overnight, in the morning 368 colonies were counted.

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! ! i) Calculate the individual dilution factors for step 1, step 2 and step 3.

ii) Calculate the total dilution factor of the bacterial cells that were plated.

iii) What is the density of viable cells in the original bacterial cell suspension (cells/ml)?

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QUESTION 2 A cell suspension has a density of 2.3 x 107 viable cells/ml. In the first step of a dilution plating experiment, 0.1 ml was added to 9.9 ml of diluent, and mixed. In the second step, 0.1 of the diluted suspension was added to 4.9 ml of diluent, and mixed. Finally, 100 μl of the final serially-diluted suspension was spread on nutrient agar plates and incubated at 37˚C overnight.

i) Draw a sketch of the dilution scheme.

ii) Calculate the total dilution factor for the cell suspension that was spread on the nutrient agar plate.

iii) What volume (in ml) of the cell suspension was plated?

iv) How many colonies would you expect to observe on the plates in the morning after incubation at 37˚C?

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QUESTION 3 Three consecutive dilutions of an E. coli bacterial cell suspension were carried out. The first was a 100 fold in sterile saline solution, followed by another 100 fold dilution and finally a 50 fold dilution. 0.1 ml of the final serially-diluted suspension was spread on a nutrient agar plate, and incubated at 37˚C overnight. In the morning, 219 colonies were counted. What was the density of viable cells in the original cell suspension (cells/ml)?

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QUESTION 4 In the first step of a dilution plating experiment, 2 ml of a suspension of bacterial cells was added to 198 ml of diluent, and mixed. In the second step, 1 ml of the diluted suspension was added to 49 ml of diluent, and mixed. Finally, 0.15 ml of the final serially-diluted suspension was spread on a nutrient agar plate, and incubated at 37˚C overnight. In the morning, 409 colonies were counted. What was the density of viable cells in the original undiluted cell suspension (cells/ml)?

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QUESTION 5 100 μl of a suspension of bacterial cells was initially added to 3.9 ml of diluent, and mixed. This was further diluted 70 fold, and then 0.2 ml of the final serially-diluted suspension was spread onto a nutrient agar plate. The plate was incubated at 25˚C for 3 days and 389 colonies were counted. What was the density of the viable cells in the original bacterial cell suspension (cells/ml)?

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QUESTION 6 A cell suspension has a density of 4.3 x 106 viable cells/ml. A dilution plating experiment was carried out to determine the number of viable cells present in the cell suspension. The protocol was as follows: 1st: 0.1 ml was added to 8.9 ml diluent, and mixed. 2nd: 0.1 of the diluted suspension was added to 4.9 ml diluent, and mixed. 3rd: 0.2 ml of the final serially-diluted suspension was spread on nutrient agar plates and incubated at 37˚C overnight.

How many colonies would you expect to observe on the plates in the morning?

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