Add 1ml bacteria to 99mls then 1m to 5 tubes of 9ml. Final and initial dilutions are required. Final dilution factor for plates are required. Plate one is from tube 3 adding 0.1 ml. Plate 2 is from tube 4 adding 1m. Plate 3 from tube 4 adding 0.1m. Plate 4 is adding 1ml ( 30 colonies are given for this plate) from tube 5. Plate 5 is adding 0.1 ml from tube 5. We are asked to give the expected number of colonies for plates 1,2,3,5. We are asked how many microbes/ml are in orginal sample of bacteria.
I know 99 ml + 1ml is 1/100 or 10 negative 2 and is the intial and final dilution for 99 ml tube, but I have had no clear instruction on the step by step procedure of the problem! Please give clear and simple instructions so I may solve this diluton problem! Thankyou!
It looks like you have been given a classic "dilution series" problem to do, where you are being asked to prepare a number of tubes containing bacterial cultures diluted in a stepwise fashion. This technique can be coupled with "plating" to determine the concentration (number of bacterial cells per milliliter) of the original culture solution used to prepare the dilutions. It is worthwhile reviewing the theory behind these technique, so I will give you the url for a website that you can refer to for details about this. In brief (from http://www.textbookofbacteriology.net/growth_2.html ):
"Indirect viable cell counts, also called plate counts, involve plating out (spreading) a sample of a culture on a nutrient agar surface. The sample or cell suspension can be diluted in a nontoxic diluent (e.g. water or saline) before plating. If plated on a suitable medium, each viable unit grows and forms a colony. Each colony that can be counted is called a colony forming unit (cfu) and the number of cfu's is related to the viable number of bacteria in the sample.
Advantages of the technique are its sensitivity (theoretically, a single cell can be detected), and it allows for inspection and positive ...
A classic "dilution series" problem is achieved.