# calculation help

### #1

Posted 30 June 2011 - 11:55 AM

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

### #2

Posted 30 June 2011 - 12:47 PM

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 ...

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

### #3

Posted 30 June 2011 - 01:10 PM

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 [ab] = 1mg/ml)?

Does that look right?

Thanks

**Edited by whatisscience?, 30 June 2011 - 01:33 PM.**

### #4

Posted 30 June 2011 - 01:16 PM

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 [ab] = 1mg/ml)?

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?

**Edited by pito, 30 June 2011 - 01:24 PM.**

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

### #5

Posted 30 June 2011 - 01:28 PM

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 [ab] = 1mg/ml)?

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

### #6

Posted 30 June 2011 - 01:39 PM

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 [ab] = 1mg/ml)?

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...

**Edited by pito, 30 June 2011 - 01:59 PM.**

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

### #7

Posted 30 June 2011 - 01:57 PM

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.

**Edited by pito, 30 June 2011 - 02:03 PM.**

### #8

Posted 01 July 2011 - 05:08 AM

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

**Edited by whatisscience?, 01 July 2011 - 05:10 AM.**

### #9

Posted 01 July 2011 - 06:44 AM

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?

### #10

Posted 05 July 2011 - 05:59 AM

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

### #11

Posted 05 July 2011 - 10:20 AM

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

Allright.

OK, there are 2 ways to solve this.

Either you start from a fixed amount, mass (or volume) of strept or you start from a fixed amount of Antibody.

I dont know which you prefer so I start with a fixed amount of antibody.

You said that you have 100µg of antibody at a concentration of 1mg/ml.

We know that tha AB is 150dalton = 150 gram/mol

So 150gram = 1mol

100µg = 6,667 x 10^-7 moles

now if we have 6,667 x 10^-7 moles of antibody, we have 6 times this amount of biotine since we have 6 biotine molecules per 1 molecule AB...

6,667 x 10^-7 moles times 6 = 4 x 10^-6 moles biotine

we know that 2.10^-9 mol biotine is 1mg strept.

So 4 x 10^-6 moles of biotine = 2000mg strep = 2gram strept.

And we also know:

Concentration of strept mag bead : 0.1% w/v

Amount of beads in vial: 2mL

which means we have 0,002 gram strept per vial so we need 1000 vials for our 100µg AB.

this is a bit too much I think.

Are you sure its 0,1 % ? because this seems very low.

Now the other method is the other way around.

Lets start with a fixed amount of strept.

Lets take 1 vial , 1 vial equals 0,002mg of strept

so we have 0,002mg strept, for 1mg strept we need 2.10^-9 mol biotine, this makes for 0,002mg a total amount of 4 x 10^-12 mol biotine

4 x 10^-9 mol biotine is 4 x 10^-12 /6 moles of AB (6Biotines for 1 AB) = 6,667 x 10^-13 moles AB

6,667 x 10^-13 moles AB with 150gram/mol

So 1 mol = 150 gram

6,667 x 10^-13 moles = 1,00005 x 10^-10 gram or 0,000100005µg AB.

and you know its 1mg/ml for the antibody so you can find how much ml this is.

Now I dont know what the objective is of this test or what you are trying to do.. but these numbers seem strage..very small...

I dont work with antibodies etc.. so I cant tell you if they are normal or not..

Check the calculations. Maybe I made an error somewhere.

### #12

Posted 24 August 2011 - 06:49 AM

Sorry for such a late response. But when I said 150,000 daltons I meant 150000 not 150 daltons (sorry for the confusion). So that would cut down our number by a factor 1000 and so we would need 1.90mg of strept = 1.90 mL of strept so we would need only 1 vial per 100ug of antibody right?

### #13

Posted 24 August 2011 - 07:45 AM

Pito,

Sorry for such a late response. But when I said 150,000 daltons I meant 150000 not 150 daltons (sorry for the confusion). So that would cut down our number by a factor 1000 and so we would need 1.90mg of strept = 1.90 mL of strept so we would need only 1 vial per 100ug of antibody right?

You just need to redo the math with 150000 in stead of 150, so yes, there should a factor 1000 difference.