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


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#1 .Bioforum.

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Posted 12 March 2011 - 06:16 PM

We transferred 0.1ml from original sample and diluted with peptone and got various dilutions. then did colony counts and got these results:

No. of bacteria / ml:

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)

BUT what was the number of bacteria in original sample??

Do I average the no of organism/ml from the 3 dilutions and would that be the total no of bacteria in the original sample?

Please help!


#2 pito

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Posted 13 March 2011 - 04:47 AM

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?
Same for your dilutions.....

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:

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?
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#3 .Bioforum.

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Posted 13 March 2011 - 05:24 PM

A 'Thank you' is such a small word for the favour you have just done...you explained me the whole thing so clearly...my tutors should quit their jobs and stay at home...i'm a final year student doing biomedical sciences and i was embarrassed that i cudn't even do this calculation :(

Yeah, I meant 10^3 on 5.28 x 10^3 and not 103 :)

So, about my plates:

Only plate 2 (45) and 1 (176) are ok.

So,
in Plate 1: there were 176 counts meaning, the no of organism/ml in original sample was= (176 x 3)/0.1= 5.28 x 10^3
in Plate 2: there were 45 counts meaning, the no of organism/ml in original sample was= (45 x 4)/0.1= 1.80 x 10^3

So, if I average 5.28 x 10^3 and 1.80 x 10^3 , I would get 3.54 x 10^3.

So am I right saying there were 3.54 x 10^3 (or 3540) bacteria in my original sample?


Once again, thanks for the detailed information..i was i had you as my tutor!

#4 pito

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Posted 14 March 2011 - 06:00 AM

A 'Thank you' is such a small word for the favour you have just done...you explained me the whole thing so clearly...my tutors should quit their jobs and stay at home...i'm a final year student doing biomedical sciences and i was embarrassed that i cudn't even do this calculation :(

Yeah, I meant 10^3 on 5.28 x 10^3 and not 103 :)

So, about my plates:

Only plate 2 (45) and 1 (176) are ok.
So,
in Plate 1: there were 176 counts meaning, the no of organism/ml in original sample was= (176 x 3)/0.1= 5.28 x 10^3
in Plate 2: there were 45 counts meaning, the no of organism/ml in original sample was= (45 x 4)/0.1= 1.80 x 10^3

So, if I average 5.28 x 10^3 and 1.80 x 10^3 , I would get 3.54 x 10^3.
So am I right saying there were 3.54 x 10^3 (or 3540) bacteria in my original sample?


Once again, thanks for the detailed information..i was i had you as my tutor!

I allready did the calculation for your plates.. check my post again.


I dont understand why you multiply with 3 or 4 ...

If you have 176 bacteria in your dilution (10^-3 or 1/1000) then you have 176 x 1000 bacteria in the non diluted sample.

I really dont understand how you calculate it with the 3 or the 4 ...?
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#5 .Bioforum.

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Posted 14 March 2011 - 11:07 AM

I am confused again.

For Plate 1, which was 10^3 dilution: I found 176 counts. So, I was told that colony forming units = (counts multiplied by dilution factor) divided by amount transferred...so here:
total counts = 176
dilution factor = 3 (from 10^3)
amount transferred in each dilution = 0.1ml

Hence, the no of organism/ml in original sample was= (176 x 3)/0.1= 5.28 x 10^3

I can see why you have multipled by 1000, which is because we made 3 serial dilution (1/10, 1/100, 1/1000)...BUT i don't get it which I should follow...

The whole thing with this exercise is that we need to compare the no of organisms in the sample over three weeks. We are incubating a sample for 3 weeks where, every week we take the sample and plate it and count....then compare the no of organisms in that sample for that week with the no of organism in the next week..so we look at the change in number of bacteria over three weeks. But I haven't been able to work out the number of organisms from these plates I have cultured!

Is colony forming units the same as number of bacteria in that sample?

I am very grateful to you as you are helping me with something that I should and need to understand and apply all my life. Really thanks a lot for this.


A 'Thank you' is such a small word for the favour you have just done...you explained me the whole thing so clearly...my tutors should quit their jobs and stay at home...i'm a final year student doing biomedical sciences and i was embarrassed that i cudn't even do this calculation :(

Yeah, I meant 10^3 on 5.28 x 10^3 and not 103 :)

So, about my plates:

Only plate 2 (45) and 1 (176) are ok.
So,
in Plate 1: there were 176 counts meaning, the no of organism/ml in original sample was= (176 x 3)/0.1= 5.28 x 10^3
in Plate 2: there were 45 counts meaning, the no of organism/ml in original sample was= (45 x 4)/0.1= 1.80 x 10^3

So, if I average 5.28 x 10^3 and 1.80 x 10^3 , I would get 3.54 x 10^3.
So am I right saying there were 3.54 x 10^3 (or 3540) bacteria in my original sample?


Once again, thanks for the detailed information..i was i had you as my tutor!

I allready did the calculation for your plates.. check my post again.


I dont understand why you multiply with 3 or 4 ...

If you have 176 bacteria in your dilution (10^-3 or 1/1000) then you have 176 x 1000 bacteria in the non diluted sample.

I really dont understand how you calculate it with the 3 or the 4 ...?



#6 mdfenko

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Posted 14 March 2011 - 12:15 PM

dilution factor = 3 (from 10^3)


10^3 does not mean 3x10. it means 10x10x10 (10 cubed, 10 to the third power, 1000). you should not multiply your bacteria count by 3 (or 4 for the next count).
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#7 pito

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Posted 15 March 2011 - 01:45 AM

mdfenko allready stated it.

I was assuming thats why you used that 3 (10^-3), but thats off course not correct.

Its very simple really: if you dilute something 1000 times you need to multiply it with 1000 ... (this is also why I wrote 1000 in my previous posts and not 10^-3, ..... using "cool" math stuff like exponents is nice, bnut you need to understand it!)

I really find this strange, didnt you learn this at school?


Anyway: its simple

the rule is: if you dilute by X, just multiply by X to know the no in the original sample.

If I take a sample and dilute it 66 times.. I multiply with 66.. same goes for 1000 or 10^-3 or 199999999
....

(and remember to also keep in mind "dilutions" you made "without diluting" , by this I mean: if you take 0,1ml from a tube that contains 1ml you allready took 1/10 of this tube, dilute it 10 times (so nr of bacteria if you plate it out is X10 in the original or if you for exammple would add this 0,1ml to a 0,9 ml volume, and then plate it out: you would have diluted it twice: 10 times dilution in step one (where you took 0,1ml from 1ml) and step 2 where you placed the 0,1 ml in 0,9 ml before plating)

Check:
http://www.accessexc...1993/serial.php
http://www.waksmanfo...er/dilution.htm
http://abacus.bates..../dilutions.html
http://www.mansfield...icrobiology.htm

maybe these websites are helpfull

and colony forming units is not really the same as bacteria.
A colony forming unit means that in that unit there was 1 bacteria capable of forming that unit (but it was possible that more bacteria just "dropped" on that same place). (but in that CFU, you have more bacteria, they have grown)


Anyway: check this link about CFU, we had a discussion about it a few months ago.
http://www.protocol-...n-microbiology/

Edited by pito, 15 March 2011 - 03:05 AM.

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