Consider the following dilution scheme:
A. 1oomL lake water
B. 5mL lake water into 45 mL diluent 0.1 ml - colonies TNTC
C. 1mL from B into 9ml diluent 0.1 ml - colonies 425
D. 0.5mL from C into 4.5 mL diluent 0.1 ml - colonies 58
E. 2mL from D into 19mL diluent 0.1 ml - colonies 12
F. 0.1mL from E into 0.9 mL diluent 0.1 ml - colonies 1
a. What dilution is achieved by adding 2mL of incoulum to 19mL of a diluent?
b. Would you expect any change in the answer of the above problem if the firat dilution was made by adding 1 mL of sample to 9 mL of diluent? Why or why not?
c. What are the total number of CFU's in the entire 100mL amount of the original lake water sample?
HELP! I'm stumped on how to do this!!!!
Dilution schemes are used to decrease the number of bacteria in a sample to a countable level. This is an example of a dilution series that will lower the bacterial numbers enough so you can count colonies on a petri plate. Each colony is assumed to come from one bacterium so from the number on the plate you can calculate how many was in your original sample. If you plate your lake water directly on a petri plate then there will be billions of bacteria and they will completely cover the plate. This does not help you determine how many cells were in the water. I will walk you through the dilution so you understand what is occurring.
In order to successfully understand dilutions you can use the following formulas.
? To find a dilution of a single tube the dilution = volume of sample/(volume of sample + volume of diluent).
? To find a total dilution after doing a dilution series using multiple tubes (this is called a serial dilution) you multiply the dilutions of the individual tubes. For instance when a sample diluted 1/10 is added to the next tube to make a 1/100 dilution the total dilution for the dilution series is 1/10 x 1/100 = 1/1000
? To find the number of bacteria/ml in the original sample after growing bacteria on a plate you use the following formula; CFU/ml = (# colonies on a plate)(1/volume used to plate bacteria)(1/dilution of that tube)
I will walk through the dilution series provided in this problem and you can see how these formulas are used to determine the dilution in each tube, the total dilution of the lake water, and the final count of bacteria.
The first dilution was to take 5 ml of lake water and put it into 50 ml of diluent. This gives you a 5/50 dilution which can be reduced down to 1/10. When you look at the 5/50 this was determine because the top number is the number of ml that you transferred and the bottom number is the number of ml total now in the tube (5ml lake water + 45 ml diluent = 50 ml total). 5/50 can then be reduced to 1/10 because 5 divides into 50 10 times. A 1/10 ...
When determining the number of bacteria in a sample you must perform a serial dilution followed by plating bacteria onto a petri plate, which is called determining the colony forming units. This explanation will walk you through the steps of a serial dilution. It explains how to do the calculations that must be performed to dilute a bacterial sample. It also explains how to calculate the number of bacteria in an original sample after the dilutions have been done and the number of bacterial colonies on each petri plate have been counted.