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:
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:
