I have taken 10 g of sample + 90 ml of sterile saline water and then it is serially diluted to 10-4. From that, 0.1 ml sample is taken in agar plate following spread plate technique and it is incubated at 37 °C for 48 hr. I have counted the number of colonies and found around 50(say). The formula is calculated as log (cfu/g) :
Number of colony/dilution × volume of sample taken × 1/weight of sample
Since 0.1 ml = 1/10ml
Number of colony/10-4 × 1/10 × 1/weight of sample
Since 10-4 = 1/104
= 50 × 104 × 10/10 = 50 × 104 = log 5 × 105 = 5.70
Or,
Number of colony × 1/dilution × volume of sample taken
Since 0.1 ml = 1/10 ml
Number of colony/10-4 × 1/10
Since 10-4 = 1/104
= 50 × 104 × 10 = 50 × 104 × 10 = log 5 × 106= 6.70
Which one I have to follow?
Hi Jag pal,
You have taken 100 ul (0.1ml) sample from your stock which you prepared by suspending 10g in 90 ml water. You have got 50 CFU/100ul which is equivalent to 500CFU/ml. I am not using any dilution factor here as you have not diluted your sample. Now you can convert your 500 CFU/ml to grams :
As you know your 90 ml water is in 10 grams
hence 1ml will be in 10/90 = 0.111g
hence your
500CFU/ml = 500 CFU/0.111g = 4545.45 CFUs/g
Hope it helps
Hi,
Following suggested procedure from Bacteriological Analytical Manual of US FDA, I will calculate the colony counts in this manner
CFU/g = (average no. of colonies x total dilution factor)/ volume plated
so, given the conditions, that will be
= (50 coloniesx 100,000)/ 0.1ml
with that, 5.0 x 10exp7 CFU/g or CFU/ml
Dilution (the reciprocal of which is the dilution factor) is dimensionless, since it is just a ratio of the parts of your sample and diluent.
to get the log CFU, then log transform this value.
log 5.0 x 10exp7 CFU/g is equal to 7.70.
http://www.fda.gov/Food/FoodScienceResearch/LaboratoryMethods/ucm063346.htm#conventional
Hope this helps.
Best regards,
Kris
Dear Harish and Kris. Thanks for your kind suggestion. Dear Kris, you have done the writing mistake in calculation. It should be 10000 not 100000. Anyway thanks. I got the idea. But I am having a little confusion in estimation of E. coli and S. aureus count. I have followed the same protocol of total plate count except changes in media. 40 colony is obtained in 10-3 dilution factor. 40 x 1000/0.1= log 4 x 105= 5.6. For total plate count count, the limit is 7 log cfu/g. What is the limit of acceptability for E. coli and S. aureus in seafood? If you have any literature or details, could you send me? thanks
Dear Jag Pal,
The cfu cannot be 5.7 or 6.7 in the higher dilutions
To calculate in very simple way please follow APHA/BAM/ISO/IS methods for enumeration of total bacteria
Just multiply the CFU (consider the dilution 20-200 colonies) with the dilution and volume of sample plated in ml
In your case 50 (colonies) x 10000 (dilution)x 0.1(volume of sample) = 50000 cfu/gram
And you cannot estimate the exact number of E.coli or S.aureus quantitatively. You can only detect the presence of those.
Hi,
in 0.1 ml you have 50 colonies
so in 1 ml u have 500 cfu
you have diluted the samples upto 10-4
so the actual sample contain 500 x 10^4 = 5 x 10^6 cfu/ml
so in 90 ml there is 90 x 5 x 10^6 = 4.5 x 10^8 bacteria
but in 90 ml, There is 10g of sample.
so cfu in 1 gm = 4.5 x 10^7
Answer : 4.5 x 10^7 cfu / gm of sample.
Sorry dear,
a small correction
Just multiply the CFU (consider the dilution 20-200 colonies) with the dilution and divide it by volume of sample plated in ml
In your case 50 (colonies) x 10000 (dilution)/ 0.1(volume of sample) = 5000000 cfu/gram of sample
You are welcome Japal!
Just search harder in PUBMED you will get plenty of references!
Good luck for research
Dear all,
May I ask how to do serial dilution to count CFU of bacteria from stock 10 ul?
Find out psb population in a gram of soil in which 29 colonies of psb is obtained in 10'5 dilution
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