Calculating molar ratio & dilution of protein cluster solution - (Jun/23/2014 )
Hi Everyone,
I'm new here and I've exhausted all other options. I'm trying to cluster a protein onto human IgG, can I cannot for the life of me get my head around the maths needed to get the molar ratio of the protein:IgG correct!
Here's what I have:
I need a 5:1 molar ratio of protein:IgG diluted to 200ng/mL (so I can add it to my culture plate)
I have:
- Protein stock = 1mg/mL = 48.6kDA = 20.58uM/mL
- IgG stock = 62mg/mL = 150kDA = 413.33uM/mL
I've done a ton of different calculations and come up with a ton of different answers ... I'm just confusing myself too much
ANY help would be greatly appreciated!!
Protein stock = 1mg/mL= 1ug/ul
Protein 1mg = 20.56nM so 0.0486mg/mL =1uM/ml
IgG stock = 62mg/mL, so 0.0161mg/mL = 1uM/mL
5:1 Molar ratio of Protein : IgG = 0.243mg/mL :0.0161mg/mL
i.e IgG = 0.243mg/mL + 0.0161mg/mL
0.2591mg/mL
N1V1= N2V2
259.1ug X V1 = 0.2ugX 1 mL
hey jess1289 and inbox, watch your units: uM=umoles/liter (nmoles/ml), uM/ml is meaningless
a mg/ml+ b mg/ml= (a+b)/2 mg/ml. i think you meant mg not mg/ml (especially since you got the units right in the final equation)
Thank you so much to both of you! Just one question Inbox ...
How did you calculate:
Protein 1mg = 20.56nM so 0.0486mg/mL =1uM/ml
IgG stock = 62mg/mL, so 0.0161mg/mL = 1uM/mL
????
@mdfenko:- Yes putting M is complete unit in itself so no need to add extra /lit or /mL. Jess, you need to check if it is umoles or uM.
@ jess
1/20.56= 0.0486
1/62 = 0.0161
Inbox, it is uM :)
Also, I understand:
Protein: 1mg = 20.56nM, therefore 1nM = 0.0486mg/mL
BUT
How does 1nM of IgG = 0.0161mg/mL if 62mg/mL = 413.33nM ??
See, I'm completely and utterly confused! Thank you so so much for the help, my experiment is at a standstill until I can work this out :(
Your first equation in the last message makes no sense.
1mg = 20.56nM is meaningless, since it sets a weight (1 mg) equal to a concentration (20.56 nM).
You are being thoroughly confused by mixing up moles with molar. The big M means molar, with units of moles/liter.
I'd strongly recommend that you write the units along with every number you use. Make sure the units match on both sides of the equation.
The abbreviation for mole is mol, not M !!
You wrote in your first message:
• Protein stock = 1mg/mL = 48.6kDA = 20.58uM/mL
• IgG stock = 62mg/mL = 150kDA = 413.33uM/mL
Now, let's take this one step at a time. You have a protein stock, with a concentration of 1 mg/ml. The protein has a molecular weight of 48.6 kDa, or (same thing) 48.6 kg/mole. From these two numbers, we can calculate the molar concentration of your protein solution. It is <1 mg/ml> / <48.6 kg/mole>. The grams cancel, leaving <1e-3 / 1e-3 liter> / <48.6 e3 / mole>. We move the mole topside, cancel the 1e-3's, and get < 1 mole/liter> / 48.6e3 = 0.00002058 mole/liter, or 20.58 uM
Similarly for the igG stock, you have 62 g/liter (cancel the milli) and 150e3 g/mole. Dividing, you have 62/150e3 mole/liter = 0.0004133 mole/liter = 413 uM.
You want to end up (at least this is how I interpret your first question) with 200 ng/ml igG protein in your sample. Since this would be very dilute, I would recommend, from a practical standpoint, that you make a solution 100x more concentrated than this, at 20 ug/ml, and plan on diluting 100x when adding it to your plate. Likely you'll need to add medium or other compounds to the plate at the same time, so this would make that easy.
So, let's make a 20 ug/ml solution of igG. You have 62 mg/ml. You need to dilute this by a factor of <62 mg/ml> / <20 ug/ml> = 3.1e3. So, if we want to make 10 ml of a 100x stock, then we need to add <10 ml>/3.1e3 of your igG stock. This is about 3.2 ul of your igG stock.
Also, into that stock, you want 5 times the molar amount of your protein. What is the molarity of your diluted igG stock? It is <20 ug/ml> / 150 kg/mole = 20e-3 / 150e3 mole/liter = .133e-6 M =133 nM.
Five times this is 667 nM. So, our 100x stock solution also needs to be 667 nM in your protein. How much is this? Well, your stock is 20.58 uM. You need to dilute by <20.58 uM> / < 667 nM> = 30.85. So, you need to add <10 ml> / 30.85 = 324 ul of your protein solution.
The remains of the 10 ml can be whatever buffer you might need, and you'll add it to the 3.2 ul of igG stock and 324 ul of your protein stock to make 10 ml final volume.
When you use it, you'll dilute this 100x before adding it to your cells.
Don't forget to filter sterilize it.
You can check this all. You've added 3.2 ul of your igG stock. This is 3.2 ul * 62 mg/ml = 200 ug in a final 10 ul volume or 20 ug/ml (check). The molarity of this solution will be <20 ug/ml> / 150 kg/mole = 133 nM (check)
You added 324 ul of your protein solution, at 1 mg/ml. This is 324 ug. This is 324e-6 g / 48.6 kg/mole = 6.7 nano moles in 10 ml of stock, for a concentration of 667 nM (check). This is five times the concentration of igG (check).
Thank you so much Phage434! It's so much clearer now!
Quick question though, is it possible to make up a 667nM solution of protein and a 133nM solution of IgG to store. Then when I want to use them add them together and make it to 20ug/mL? For this experiment, the protein and IgG need to sit and 'cluster' for an hour before I add it to my culture, and it needs to be clustered fresh each day.
Thanks again :)
phage434 on Wed Jun 25 02:42:55 2014 said:
This is 3.2 ul * 62 mg/ml = 200 ug in a final 10 ul volume or 20 ug/ml (check). The molarity of this solution will be <20 ug/ml> / 150 kg/mole = 133 nM (check)
i believe you meant "ul".
very nice explanation, by the way.
Sorry for calculation mistake.
62/413.33 = 0.150mg/mL = 1uM
5:1 Molar ratio of Protein : IgG = 0.750mg/mL :0.0161mg/mL
0.7661mg/mL
can dilute as per need of final concentration in medium.