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Calculating concentrations of Antibodies when mixing


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#1 sarasbluegroove

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Posted 26 May 2010 - 11:09 PM

This may sound like a stupid question, but...

I need to mix a detection antibody, an antigen, and another control antibody. I want them to each to be a specific final concentration in a solution. What makes sense to me is to treat them like mixing seperate chemicals. I am confused because i know each antibody will dilute the others (2 are lower concentrations, one is much higher) Originally i used this equation and mixed my solutions 2 at a time:
(C1V1)+(C2V2)=CfVf
I don't think this is right, because my ratios that i want the detection antibody and antigen at were not right. I want my concentrations to be:

.01 ug/uL of antigen
.0001 ug/uL detection antibody (of the above antigen)
.0002 ug/uL of another antibody that has nothing to do with the first 2 solutions (this is my control)

all in 1 solution.
I found this article, that talked about mixing antibodies, and used this equation to figure out the initial volumes of each to add to make a certain final concentration for each. It involved making 2 equations and using them to solve for V1 and V2. (my V2 is mult by 2 because 2 of the antibodies are the same concentration)

C1V1=C2(Vc+V1+2V2)
Vc=the total concentration of liquid wanted, so i want a solution that is 300uL, so i have 300 of water, then i add the antibodies to that.
Does this sound right? How would you do it? Am i thinking about this way too hard? Thank you.

#2 leelee

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Posted 26 May 2010 - 11:57 PM

Hmmmmm I wouldn't figure it out like that at all!! I am not a maths person so I use as few formulas as possible!!

I would decide on a final volume (say 300ul) then figure out how much of each component I would need individually using c1v1=c2v2.

So say you figured out you would need 5ul of Ag, 10ul of Ab and 20ul of control Ab.
Then add 265ul water to a tube and add the above for a final volume of 300ul.

Does that make sense??

Edited by leelee, 26 May 2010 - 11:58 PM.


#3 sarasbluegroove

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Posted 27 May 2010 - 12:25 AM

Yes, that's what i thought at first, but that doesn't take into consideration how much the lower concentrations will dilute the other higher concentration.
Here are my concentrations:
Det-ab intitial= .01 g/L
Antigen initial= .03 g/L
Ab-control initial= .02 g/L
I want these final concentrations each, but they will be mixed together:
Det-ab final=.0001 g/L
Antigen final= .01 g/L
Ab-control= .0001 g/L

If i calculate them each seperatly, it wouldn't take into consideration the others being added, it would just be considering the whole volume. i don't know, i've been thinking about this for so long i don't know which direction is up or down, lol. Thanks!

#4 Gerard

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Posted 27 May 2010 - 12:56 AM

To make 1000 ul mixed solution you should add
Det-ab 100 times diluted you need 10ul
Antigen 3 times diluted you need 333 ul
Ab-control 200 times diluted you need 5 ul
Diluent you need 1000-(10+333+5) ul

This way you diluted to 1000 ul and take all the added volumes in account.
Ockham's razor
Pluralitas non est ponenda sine necessitate
-- "You must assume no plural without necessity".

#5 Prep!

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Posted 27 May 2010 - 01:00 AM

Yes, that's what i thought at first, but that doesn't take into consideration how much the lower concentrations will dilute the other higher concentration.
Here are my concentrations:
Det-ab intitial= .01 g/L
Antigen initial= .03 g/L
Ab-control initial= .02 g/L
I want these final concentrations each, but they will be mixed together:
Det-ab final=.0001 g/L
Antigen final= .01 g/L
Ab-control= .0001 g/L

If i calculate them each seperatly, it wouldn't take into consideration the others being added, it would just be considering the whole volume. i don't know, i've been thinking about this for so long i don't know which direction is up or down, lol. Thanks!



i think u are too cunfused.. why wud one dilute the other wen each is in the specified final concentration!!!

in your eg. i would take 10 mcl of Det-Ab 333.33 mcl of Antigen and 5 mcl of Ab-control and make up the volume for a 1 liter solution!!!

think of it like this.. if i want to prepare 50% methanol 10% acetic acid 40% water solution to fixx a polyacrylamide ge. I would just measure volumes of each component and make up to the desired volume.. none of the compound will change any dilution of any other compound!!!

Edited by Prep!, 27 May 2010 - 01:03 AM.

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Cheers!!!

#6 sarasbluegroove

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Posted 27 May 2010 - 01:30 AM

Thanks everyone, i wish i had found this forum a few weeks ago! I originally calculated what you guys said, but was doubtful it was that easy because my boss said something to me that made me think it was harder. Sometimes my own brain gets in the way. Thanks again!

#7 Prep!

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Posted 27 May 2010 - 01:44 AM

Thanks everyone, i wish i had found this forum a few weeks ago! I originally calculated what you guys said, but was doubtful it was that easy because my boss said something to me that made me think it was harder. Sometimes my own brain gets in the way. Thanks again!



yeah sometimes our brain runs too fast... but then tat is why we are scientists right!!! ;) :P
Support bacteria - They are the only culture some people have!!!
Cheers!!!




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