If one is spotting the 20 microlitre spots directly and dilutions from total 1 ml of unknown CFU sample. If one gets one or two colonies on two plate and no colonies on plate on other four plates. How one can interpret the data on categorised scatter plot. Specifically question are below
1) how to take average or mean ?
2) How to show the points of clear plates ?
If one colony appears it has log value of 1.698. But if no colony it may be any value between 0-1.698.
The method you describe seems to be a variation of the Miles & Misra (1939) method which works best when the CFU/ml is 10^3 or higher. In your case you can use the MPN technique as your counts are low.
I am unclear of Dr. Singh's methodology here. What dilution is he referring too? Perhaps he should plate undiluted or less diluted samples to get a better result. With his distribution of colonies, his result could only be expressed as +-. rather than any type of statistical number or graphical result.. Joel Unowsky, Ph.D. FDA retired
In terms of determining average values, two common practices are to set values below the detection limit at half the detection limit, or alternatively to the detection limit over the square root of 2. So in your case, with a detection limit where 1 colony actually represents 50, setting values at half the detection limit, 25, would correspond to log cfu of 1.4. Of course you should indicate your calculation methodology.
He should plate 0.1 or 0.25 ml if his solution is undiluted and would get some numbers that make sense. With these small numbers and blank plates, I think + - is the best statement of his findings. Since there are 4 blank plates, one colony does not represent 50. It would represent perhaps, 2/6 x50 which is pushing it standard deviation analysis here is nonsense and A mean certainly makes no sense here nor does a median or mode with these kinds of numbers. Certainly not robust. Does he have 1 or 2 colonies on each plate? Still unclear of his dilution, I assume the solution is undiluted.
Dear Researcher @ Dr. Eric and Dr. Jeol
I have homogenised lung in 1ml after the treatment of mice for 28 days and wish to know the bacterial load remained. In case of controls (untreated) it remains 6-7 log. However treated groups have the very low CFU remained. 20 ul of this homogenate respresents the log CFU by multiplication with 50 (in per ml or per lung). When colonies appaers that is easy to represent. But on clear plates is there a way other than spotting higher volume of homogenate ?
I am attaching a reference file for better understanding. Those encircled in red are clear plates. Although i have given them the the lowest Log CFU value equivalent to 1.698. Now my question is open for discussion.
Did you start out with this inoculum and it did not increase for 28 days in the controls-very unusual? How were the mice inoculated and treated? What antibiotic was used?
Early control (4.5 log average) was inoculated and after 28 days late control (6 log) as shown in attachment. Microbe is M. tuberculosis. It was one of AntiTb drug details of which can't be revealed.
Dear Dr Singh,
Spot plating has this one as a disadvantage. When u r spot plating the spot retains some compound too along with the bacteria which may inhibit the growth. Please try to spread 200 microliters of the sample on a plate and check your result.
I agree, spreading is safer as far as diluting the compound. The attachment s not visible to me. Where were the mice inoculated and treated, subcutaneous, intraperitoneal etc. Am surprised that all mice survive. Treatment of mice for 28 days is very unusual, the trauma of so much treatment should have killed a good percent of the mice.
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