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# .Bioforum.

Member Since 22 Dec 2010
Offline Last Active May 15 2011 08:36 AM

### #103441CALCULATING no of bacteria in original sample...after dilutions? [Very SHORT Que

Posted on 13 March 2011 - 04:47 AM

.Bioforum., on 12 March 2011 - 06:16 PM, said:

in dilution 10^-5 = 3 (so no of organism/ml is (3 x 5)/0.1= 1.50 x 102)

could you explain what you do?

I dont understand what you mean by (3x5)/0,1 = 1.5 x 102

3X5 = 15, divided by 0,1 = 150  its not the same as 1.5x102 ? I dont understand that.. Same for the other dilutions..
(I understand you divide by 0,1 because its 0,1 ml, but what you then do.. I dont get it? Maybe you made a mistake? Because dividing by 0,1 is the same as multiplying with 10... you did the opposite, but then the 102 ? Is that 10^2 ?)

Anyway:

if you have a sample that you diluted 10 times and you count 20 bacteria then how many do you have in the not diluted sample?

Now we come to the second part: what if I have more then 1 dilution (like you) then how do I solve it?

What do you think?

Example:

I have a sample , I dilute it 10 times, 100 times and 1000 times.

My 1/10 sample gives me: 100 bacteria
my 1/100 sample gives me 8 bacteria
and my 1/1000 sample gives me 2 bacteria.

First thing you do is "go back to the orginal sample"
==>
the 1/10 sample is diluted 10 times, so the original is: 10 bacteria X 10 (diluted 10 times) = 1000 bacteria
The 1/100 sample is diluted 100 times, original is thus: 8 x 100 = 800 bacteria
The 1/1000 sample is diluted 1000 times, original is thus: 2x 1000 = 2000 bacteria

Now how many bacteria were there in the original one?

Here is all depends on the rules you learned.

In general people will only use those plates that has a number of bacteria on it between 30 and 200-400.

So in this example we only use plate number 1 (the 1/10) with the 100 bacteria.

If you had more plates with bacteria in the range of 30-400 you had to take the average... (or depending on the rules , the most diluted ones or the least diluted ones)

Now, going back to your example:

Quote

in dilution 10-^3 = 176 (so no of organism/ml is (176 x 3)/0.1= 5.28 x 103)
in dilution 10^-4 = 45 (so no of organism/ml is (45 x 4)/0.1= 1.80 x 103)
in dilution 10^-5 = 3 (so no of organism/ml is (3 x 5)/0.1= 1.50 x 102)
Only plate 2 (45) and 1 (176) are ok.

Plate 1 means a dilution of 1/1000 thus 176 in the original sample means 176000 (I hope the fact that you used 0,1ml is allready calculated in the 10^-3 dilution?)

Plate 2 means a dilution of 1/10000 thus 45 would be: 450000 bacteria.

If you look at these numbers 450.000 and 176.000 then you would easly see that this is rather strange, but you will just need to take the average of both: 450.000 + 176.000 = 626.000 and divided by 2= 313.000 bacteria in the original sample.

Or, what other might do: they only use dilution 2 because thats the most diluted one.. Or other even might say: no we only use the first dilution (1/10)....

Now normally in the lab we would make more then 1 plate so extreme values would not matter so much... In your case: you only have 3 plates and those are 3 different dilutions.. So its a bit hard to really use those values.

Dont you have any rules ? Didnt your teacher tell you how to do it in the end? Because it seems weird to use this as a an example without telling you more about the rules?