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Serial dilution with a factor of 1.07E + 8 - how to?


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#1 Bug Man

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Posted 22 May 2011 - 06:01 PM

Hello Friends,

I'm a Microbiologist working in Australia - I've been in the industry for 6 years now but have never really had to work with obtaining fixed dilution bacterial cultures so I've never really had to refine my serial dilution knowledge... so the situation is this. I've fermented a stock culture and obtained a final concentration of 1.07E + 15 I have a target concentration of 1.00E + 07. What is the calculation process/working method for diluting the initial 1.07E + 15 concentratoion culture to 1.00E + 07. Obviously I can't do it in one step as 1mL in 107,000,000ml isn't practical. I was thinking something along the lines of 10uL in a total volume of 10.7mL for a 1000 fold dilution, then a further 10uL of that in to 10.7mL for a further 1000 fold dilution and then 1mL of that final dilution in to 107mL for a 100 fold dilution which should, if i've got it figured out correctly, give me 1,000 X 1,000 X 100 = 100,000,000

Any help would be appreciated

Kind regards

#2 lsek

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Posted 23 May 2011 - 07:59 AM

Hi,

I would do a 100,000,000 dilution in serial as you suggested. 10 ul of stock to 10 ml of medium (1000x diluted), 10 ul of 1000x diluted solution to 10 ml of medium (1,000,000x diluted), and 1 ml of 1,000,000x diluted solution to 100 ml medium (100,000,000 diluted).

If you want to be very precise, the actual dilution is 107,000,000. To do that, 9.35 ul stock to 10 ml medium (1070x diluted), 10 ul of 1070x dilted solution to 10 ml medium (1,070,000 diluted), 1 ml of 1,070,000 diluted solution to 100 ml medium (107,000,000 diluted).

Hope this helps

Bests,

...-...

===><===


#3 Bug Man

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Posted 23 May 2011 - 03:26 PM

Hi,

I would do a 100,000,000 dilution in serial as you suggested. 10 ul of stock to 10 ml of medium (1000x diluted), 10 ul of 1000x diluted solution to 10 ml of medium (1,000,000x diluted), and 1 ml of 1,000,000x diluted solution to 100 ml medium (100,000,000 diluted).

If you want to be very precise, the actual dilution is 107,000,000. To do that, 9.35 ul stock to 10 ml medium (1070x diluted), 10 ul of 1070x dilted solution to 10 ml medium (1,070,000 diluted), 1 ml of 1,070,000 diluted solution to 100 ml medium (107,000,000 diluted).

Hope this helps

Bests,

...-...


Thanks Isek, I appreciate your reply... is there an easy way of working out the dilutions? i.e. how did you determine the 9.35uL stock to 10mL medium for the 1070X dilution? is there an equation to use? I've actually re-worked my target concentrations and the dilution factor is now 'only' 38,909,091 - or, for the sake of ease, 39,000,000. I'm just trying to get my head around this so i really appreciate everyones assistance...

#4 Bug Man

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Posted 23 May 2011 - 05:07 PM

Isek, would you mind casting your eye over my calculation just to make sure that I have it:

I've modified the target concentration of that particular species to 2.75E+07 - the original concentration is 1.07E+15. That requires a dilution factor of 38,909,091. To reach that target i've calculated the following serial dilutions:

40 Fold: 25uL in 1mL medium
1,000 Fold: 20uL in 20mL medium
1,000 Fold: 20uL in 20mL medium

Giving me a 40*1000*1000 fold dilution or 40,000,000

This would also give me a final working volume of 20mL of stock solution

I've got 11 other species to do the same dilutions to (albeit different original concentrations):

For the bacterial species i've got a target of 2.75E+07

Bacteria 1: Initial concentration - 1.20E+11

4.37 Fold Dilution: 0.229uL in 1ml medium
1,000 Fold Dilution: 20uL in 20mL medium
Dilution Factor: approx 4,367

Bacteria 2: Initial Concentration - 1.06E+12

40 Fold Dilution: 25uL in 1ml medium
1,000 Fold Dilution: 20uL in 20mL medium
Dilution Factor: approx 40,000

Bacteria 3: Initial Concentration - 2.26E+12

83.33 Fold Dilution: 12uL in 1ml medium
1,000 Fold Dilution: 20uL in 20mL medium
Dilution Factor: approx 83,333

Bacteria 4: Initial Concentration - 7.00E+10

2.54 Fold Dilution: 393uL in 1ml medium
1,000 Fold Dilution: 20uL in 20mL medium
Dilution Factor: approx 2,545

Bacteria 5: Initial Concentration - is the above example of 1.07E+15

Bacteria 6: Initial Concentration - 8.90E+09

3.25 Fold Dilution: 308uL in 1ml medium
100 Fold Dilution: 200uL in 20mL medium
Dilution Factor: approx 325

Mould & Yeast have a target concentration of 2.75E+06

Yeast 1: Initial Concentration - 7.40E+10

27.03 Fold Dilution: 37uL in 1ml medium
100 Fold Dilution: 20uL in 20mL medium
Dilution Factor: approx 27027

Mold 1: Initial Concentration - 4.40E+07

16 Fold Dilution: 1mL in 16ml medium
Dilution Factor: 16

Yeast 2: Initial Concentration - 4.70E+08

27.03 Fold Dilution: 584uL in 1ml medium
100 Fold Dilution: 20uL in 20mL medium
Dilution Factor: approx 171

Mold 2: Initial Concentration - 6.50E+08

2.36 Fold Dilution: 423uL in 1ml medium
100 Fold Dilution: 200uL in 20mL medium
Dilution Factor: 236

Mold 3: Initial Concentration - 1.40E+06

No dilution necessary

Mold 4: Initial Concentration - 1.02E+11

2.36 Fold Dilution: 27uL in 1ml medium
100 Fold Dilution: 20uL in 20mL medium
Dilution Factor: 37037


I'm sorry for putting all of these up here - i'm hoping though that it might be useful for ones who read this in the future to figure out what i've done...

Any help/advice is appreciated

Kind regards

Edited by Bug Man, 23 May 2011 - 05:09 PM.


#5 pito

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Posted 24 May 2011 - 01:52 AM

I have not checked them.

But I would give you the following advise to start with: put all the units in the same unit.

I mean: 1Ál in 1ml is a dilution, but how much? Its easier if you said: 1Ál in 1000Ál, you see my point?

its possible you make stupid mistakes due to units..
I am not saying you did make mistakes, but its clear that you have doubts when working with dilutions.

Lets take an example:

For the bacterial species i've got a target of 2.75E+07

Bacteria 1: Initial concentration - 1.20E+11

4.37 Fold Dilution: 0.229uL in 1ml medium
1,000 Fold Dilution: 20uL in 20mL medium
Dilution Factor: approx 4,367


Initial concentration is 1.2 x 10^11 and you need to get it to 2.75 x 10^7

So, 2 things are clear: you need to go from 1.2 to 2.75 and from 10^11 to 10^7.

First we go from 1.2 to 2.75
1.2 x 10^11 is the same as 12 x 10^10

So how do we go from 12 x 10^10 to 2.75 x 10^10 ?
as you said: 4.37 dilution.

So we take 1ml of your 12 x 10^10 and add that in a total volume of 4,37 ml (add 3,37ml water).
(check your units, you say you dilute 4.37, but you work with Ál and ml... is it correct what you do? 0,229Ál in 1 ml, is that really a dilution of 4,37?)

And after this step its easy to make the other dilutions, because thats only dilution 10 times or 1000, times...


I havent got time to check all your example, but start first by checking your units again.

ANd make easy dilutions, its easier to take 1 ml of your stock and add X ml to dilute 6,7 times for example then to work with 0,87 ml of your stock and add 1ml.. its easier to calculate... try it yourself.

(what would be easier: 1 ml of stock with 2,5ml , to get 3,5 dilution or 0,4 ml in 1ml?)

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.


#6 Bug Man

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Posted 24 May 2011 - 03:55 PM

So we take 1ml of your 12 x 10^10 and add that in a total volume of 4,37 ml (add 3,37ml water).
(check your units, you say you dilute 4.37, but you work with Ál and ml... is it correct what you do? 0,229Ál in 1 ml, is that really a dilution of 4,37?)


Thanks Pito for your reply... when i was doing the actual calculations i worked in mL so 229uL was 0.229mL

In regard determining the culture volume i simply decided on my final volume and divided that by the dilution factor i.e. if i wanted a final volume of 1mL and to obtain that I utilised 229uL of culture then i would add 771uL of diluent...

Does that all sound ok?

#7 pito

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Posted 25 May 2011 - 01:25 AM


So we take 1ml of your 12 x 10^10 and add that in a total volume of 4,37 ml (add 3,37ml water).
(check your units, you say you dilute 4.37, but you work with Ál and ml... is it correct what you do? 0,229Ál in 1 ml, is that really a dilution of 4,37?)


Thanks Pito for your reply... when i was doing the actual calculations i worked in mL so 229uL was 0.229mL

In regard determining the culture volume i simply decided on my final volume and divided that by the dilution factor i.e. if i wanted a final volume of 1mL and to obtain that I utilised 229uL of culture then i would add 771uL of diluent...

Does that all sound ok?


Yes sounds ok.


I understand that sometimes you need a fixed end volume... but even then you can to what I did.

Ex: if you need an end volume of 1ml, and if you prepare and end volume of 3,5 ml for example, you can still take 1ml out of that 3,5 ml.. and the advantage then is that you are SURE you have 1ml to take.. because if you prepare only 1 ml (and you need one) its always possible you end up with less.
(now in this problems its not so much a big deal, but if you work with low amounts of chemical analyses then its a big deal..)
Another reason is that you can easly multiply by 4,76 (for example) but if you dividing by that is harder.. you will need to round up more and make a bigger mistake.. but again, thats also more for very analytic things.



Just do it as you like, but my advise could help you to understand it and you can always change the volumes etc...when you understand it completely.

The most easy rule to get it is the following:

1 ml (X) + 9 ml (Y) = 10 ml (Z) (with X is stock and Y is dilutant, water for ex, Z= total volume) ====> Z is then always the dilution factor , but for this you need to have X as 1!! ... Thats why using 1 ml from stock (or 1Ál or..) is easy to see what the dilution factor is.

(for ex. 0,5 ml (X) + 4,5 ml (Y)= 5ml (Z), here you can say that the dilution (Z) was 5 .. the dilution was also 10.... )

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.





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