# calculation help - see topic (Jun/30/2011 )

We have some streptavidin magnetic beats (~0.2 um diameter) and we want to bind them onto biotin labeled antibodies, where each ab has ~6 molecules of biotin avail.

The instructions says the binding capacity of the strep beads: 2.10nmole of Biotin-molecule to 1 mg of strept. particle.

So how would I go about calculating how much antibody (mg) we need to use?

Concentration of strept mag bead : 0.1% w/v

Amount of beads in vial: 2mL

Antibody concentration: 100ug (1mg/ml)

Thanks

Well look at what you know.

You know that 1 vial contains 2ml and the concentration is 0,1%w/v. So what does this tell you?

It tells you that 2ml contains 0,1% (mass) of the strept. beads.

so you just need to take 0,1% of 2ml to know (in mass) how "much" strept you have.

(0,1% of 2ml is thus 0,002g, assuming that 1ml is 1g)

So now you know how much gram strept you have in each vial and you know the antibody cocentration (thus also how many/much antibodies...) so its not that hard to make the last calculation.

(its the same as for the strept... )

The only thing that is left for you is to see what the molar mass of biotin is..because your cocentration is given in gram and you need to know the number of moles ...

pito on Thu Jun 30 20:47:53 2011 said:

Well look at what you know.

You know that 1 vial contains 2ml and the concentration is 0,1%w/v. So what does this tell you?

It tells you that 2ml contains 0,1% (mass) of the strept. beads.

so you just need to take 0,1% of 2ml to know (in mass) how "much" strept you have.

(0,1% of 2ml is thus 0,002g, assuming that 1ml is 1g)

So now you know how much gram strept you have in each vial and you know the antibody cocentration (thus also how many/much antibodies...) so its not that hard to make the last calculation.

(its the same as for the strept... )

The only thing that is left for you is to see what the molar mass of biotin is..because your cocentration is given in gram and you need to know the number of moles ...

So the molar mass of biotin is 244 g/mol

I need 2.10e-9 mol biotin for each 1mg of strept particle which = 2.1e-9mol * 244 g/mol = 5.12e-7 g Biotin molecule for each 1mg of strept particle? which = 5.34 e-4 mL of antibody (since

Does that look right?

Thanks

whatisscience? on Thu Jun 30 21:10:14 2011 said:

pito on Thu Jun 30 20:47:53 2011 said:

Well look at what you know.

You know that 1 vial contains 2ml and the concentration is 0,1%w/v. So what does this tell you?

It tells you that 2ml contains 0,1% (mass) of the strept. beads.

so you just need to take 0,1% of 2ml to know (in mass) how "much" strept you have.

(0,1% of 2ml is thus 0,002g, assuming that 1ml is 1g)

So now you know how much gram strept you have in each vial and you know the antibody cocentration (thus also how many/much antibodies...) so its not that hard to make the last calculation.

(its the same as for the strept... )

The only thing that is left for you is to see what the molar mass of biotin is..because your cocentration is given in gram and you need to know the number of moles ...

So the molar mass of biotin is 244 g/mol

I need 2.19e-9 mol biotin for each 1mg of strept particle which = 2.19e-9mol * 244 g/mol = 5.34e-7 g Biotin molecule for each 1mg of strept particle? which = 5.34 e-4 mL of antibody (since

Does that look right?

Thanks

I dont see the factor 6 anywhere... so check it again...

(remember that you wrote: where each ab has ~6 molecules of biotin avaible... so check it...)

your calculation on itself is correct, however: you need to remember that factor 6 !

What you did is correct, but you calculated the amount antibodys, wich is correct, but every antiby has 6 molecules of biotin ...

so how are you going to fix this?

pito on Thu Jun 30 21:16:31 2011 said:

whatisscience? on Thu Jun 30 21:10:14 2011 said:

pito on Thu Jun 30 20:47:53 2011 said:

Well look at what you know.

You know that 1 vial contains 2ml and the concentration is 0,1%w/v. So what does this tell you?

It tells you that 2ml contains 0,1% (mass) of the strept. beads.

so you just need to take 0,1% of 2ml to know (in mass) how "much" strept you have.

(0,1% of 2ml is thus 0,002g, assuming that 1ml is 1g)

So now you know how much gram strept you have in each vial and you know the antibody cocentration (thus also how many/much antibodies...) so its not that hard to make the last calculation.

(its the same as for the strept... )

The only thing that is left for you is to see what the molar mass of biotin is..because your cocentration is given in gram and you need to know the number of moles ...

So the molar mass of biotin is 244 g/mol

I need 2.19e-9 mol biotin for each 1mg of strept particle which = 2.19e-9mol * 244 g/mol = 5.34e-7 g Biotin molecule for each 1mg of strept particle? which = 5.34 e-4 mL of antibody (since

Does that look right?

Thanks

I dont see the factor 6 anywhere... so check it again...

(remember that you wrote: where each ab has ~6 molecules of biotin avaible... so check it...)

your calculation on itself is correct, however: you need to remember that factor 6 !

What you did is correct, but you calculated the amount antibodys, wich is correct, but every antiby has 6 molecules of biotin ...

so how are you going to fix this?

ahh yeah i see... so 6 molecule of biotin * <244g/mol / 1molecule of biotin> * 2.1e-9mol Biotin = 2.93e-6 g of biotin

so basically multiply it by 6

whatisscience? on Thu Jun 30 21:28:47 2011 said:

pito on Thu Jun 30 21:16:31 2011 said:

whatisscience? on Thu Jun 30 21:10:14 2011 said:

pito on Thu Jun 30 20:47:53 2011 said:

Well look at what you know.

You know that 1 vial contains 2ml and the concentration is 0,1%w/v. So what does this tell you?

It tells you that 2ml contains 0,1% (mass) of the strept. beads.

so you just need to take 0,1% of 2ml to know (in mass) how "much" strept you have.

(0,1% of 2ml is thus 0,002g, assuming that 1ml is 1g)

So now you know how much gram strept you have in each vial and you know the antibody cocentration (thus also how many/much antibodies...) so its not that hard to make the last calculation.

(its the same as for the strept... )

The only thing that is left for you is to see what the molar mass of biotin is..because your cocentration is given in gram and you need to know the number of moles ...

So the molar mass of biotin is 244 g/mol

I need 2.19e-9 mol biotin for each 1mg of strept particle which = 2.19e-9mol * 244 g/mol = 5.34e-7 g Biotin molecule for each 1mg of strept particle? which = 5.34 e-4 mL of antibody (since

Does that look right?

Thanks

I dont see the factor 6 anywhere... so check it again...

(remember that you wrote: where each ab has ~6 molecules of biotin avaible... so check it...)

your calculation on itself is correct, however: you need to remember that factor 6 !

What you did is correct, but you calculated the amount antibodys, wich is correct, but every antiby has 6 molecules of biotin ...

so how are you going to fix this?

ahh yeah i see... so 6 molecule of biotin * <244g/mol / 1molecule of biotin> * 2.1e-9mol Biotin = 2.93e-6 g of biotin

so basically multiply it by 6

dont you mean divide by 6 ? Since each antibody holds 6 molecules of biotin?

You calculated the amount of biotin (thinkin 1 antibody= 1 binding space/biotin) , but in reality each antibody holds 6 binding spaces...

(its late here.. I might be wrong, I'll have to check it again tomorrow or maybe someone else will check in before I do).

Just one final remark: the concentration of the antibody, what concentration is it exactly ? Because I assumed its the concentration of biotin itself... or biotin included with the antibody because if its not.. we cant calculate it like we did.

You calculated the amount of biotin (the mass of biotin), but we need to change this in the mass of the antibody if the concentration is given for the antibody itself...

just a quick example to show what I mean, I only changed the numbers to make it easier, but you can redo it with the correct numbers:

1mg strep requires 2 moles of biotin, knowing that M of biotin is 244g/mol, this means we need 488 gram of biotin but now we need to know how much antibody this is.. and here comes the question about that concentration. Does it say Antibody concentration: 100ug (1mg/ml)as in: 1mg biotin for each ml or is it just the antibody itself or the antibody + the biotin?

I cant imagine its just the antibody itself.. but I dont work with antibodies, so I dont know how they do it...

check this.

pito on Thu Jun 30 21:57:23 2011 said:

just a quick example to show what I mean, I only changed the numbers to make it easier, but you can redo it with the correct numbers:

1mg strep requires 2 moles of biotin, knowing that M of biotin is 244g/mol, this means we need 488 gram of biotin but now we need to know how much antibody this is.. and here comes the question about that concentration. Does it say Antibody concentration: 100ug (1mg/ml)as in: 1mg biotin for each ml or is it just the antibody itself or the antibody + the biotin?

I cant imagine its just the antibody itself.. but I dont work with antibodies, so I dont know how they do it...

check this.

its the antibody concentration 1mg/ml. But there are usually 3-6 biotin molecules per antibody

whatisscience? on Fri Jul 1 13:08:47 2011 said:

pito on Thu Jun 30 21:57:23 2011 said:

just a quick example to show what I mean, I only changed the numbers to make it easier, but you can redo it with the correct numbers:

1mg strep requires 2 moles of biotin, knowing that M of biotin is 244g/mol, this means we need 488 gram of biotin but now we need to know how much antibody this is.. and here comes the question about that concentration. Does it say Antibody concentration: 100ug (1mg/ml)as in: 1mg biotin for each ml or is it just the antibody itself or the antibody + the biotin?

I cant imagine its just the antibody itself.. but I dont work with antibodies, so I dont know how they do it...

check this.

its the antibody concentration 1mg/ml. But there are usually 3-6 biotin molecules per antibody

Eum, I havent really worked with antibodies, so maybe there is a general rule to calculate it then?

But you do understand my problem with the calculation?

If you calculate the amount of biotin , you still dont know how to relate it to the antibody.. this is where I get stuck.. since I dont work with antibodies.

Dont you know anything else from the antibody? molecular mass or?

pito on Fri Jul 1 14:44:18 2011 said:

whatisscience? on Fri Jul 1 13:08:47 2011 said:

pito on Thu Jun 30 21:57:23 2011 said:

just a quick example to show what I mean, I only changed the numbers to make it easier, but you can redo it with the correct numbers:

1mg strep requires 2 moles of biotin, knowing that M of biotin is 244g/mol, this means we need 488 gram of biotin but now we need to know how much antibody this is.. and here comes the question about that concentration. Does it say Antibody concentration: 100ug (1mg/ml)as in: 1mg biotin for each ml or is it just the antibody itself or the antibody + the biotin?

I cant imagine its just the antibody itself.. but I dont work with antibodies, so I dont know how they do it...

check this.

its the antibody concentration 1mg/ml. But there are usually 3-6 biotin molecules per antibody

Eum, I havent really worked with antibodies, so maybe there is a general rule to calculate it then?

But you do understand my problem with the calculation?

If you calculate the amount of biotin , you still dont know how to relate it to the antibody.. this is where I get stuck.. since I dont work with antibodies.

Dont you know anything else from the antibody? molecular mass or?

molecular mass of ab is 150,000 Daltons