Hello ,
I 'm having the total count of 1.75*10^13 CFU/mL of bacteria. For the assay, I need 5*10^5 to 1*10^6 CFU/mL.
To understand the total dilution factor which I shall use for attaining the CFU/mL mentioned above done via the following formula
TOTAL DILUTION FACTOR = NUMBER OF COLONIES/ML PLATED/ CFU/ML
Example
5*10^5/1.75*10^13 =0.6*10^-7
My question is how to attain this 0.6
Is it representing the amount of the inoculum that I should use for dilution?
Kindly find the attachment that I used for calculation
That's what i'm confused about if it's 1*10^-7 means we have to take 10^-7 dilution that is clear.My doubt is 0.6*10^-7 dilution means what ? 0.6 representing what ?
Hi,
Daria Eroshenko
I'm so glad about your explanation but I would like to know on which basis you are taking 0.2 ml in 6.8 ml in first case and 0.4 ml in 6.6 ml in my experiment case(Total volume is 7ml).How could you conclude like 0.057 cells will be there in it.
Your valuable explanation means alot.
Hello ,
Daria Eroshenko
I'm very thankful for your valuable time and patience
It was such a wonderful explanation,with your permission I would like clarify some more doubts .
If in case i'm desiring to have a 5*10^5 CFU/ml as per your explanatory calculation if i do
1.75*10^13/5*10^5=0.35*10^8 or 3.5*10^7,
. 0.35*10 ^8/ (10 2 *10 2 * 10 2* 10^2) = 0.35mL
0.4*0.35=14
FV=14-0.4= 13.6 mL
Is this correct ?
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