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3. A sample of cell suspension is ready for counting and viability assessment; however, after loading the hemocytometer and microscopically observing the cells on the counting grids, it is determined that the counting ranges are too high (i.e. there are too many cells to count). What must be done to obtain good counting ranges? What volume of medium must be added to 0.1 ml of the original cell suspension to obtain a 1:10 dilution? What is the dilution factor? If a sample of this 1:10 dilution is added to an equal volume of trypan blue solution, what is the final dilution factor?

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Reid Wolff
Reid WolffLv2
28 Sep 2019

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Related questions

Two T-75 flasks containing anchorage-dependent cultures were trypsinized. The cells from both flasks were pooled, and the resulting cell suspension was transferred to a conical tube. The total volume of liquid cells suspended in media was 20.5 mls. Although these cells are routinely split 1:2, anyone handling these cultures is requested to assess the viability of the population. 0.5 ml of the cell suspension was added to a tube containing an equal volume of trypan blue, the tube was agitated and hemocytometer was loaded with the resulting suspension. Both grids were covered with appropriate amount of cell suspension, and the following counts were obtained. Grid #1: 5 squares counted for a total of 176 cells (5 out of 176 cells were stained blue) Grid #2: 5 squares counted for a total of 195 cells. (5 out of 195 cells were stained blue) What is the average number of viable cell per square? What is the % viability? What is the total number of cells available for subculturing? Since these cells are routinely split 1:2, approximately how many millions of cells will be used to seed each of the 4 flasks?

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:

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