In principle, assuming you use the conditions specified for the reaction and the substrate you are using is a good one for the enzyme, you can calculate how much enzyme to use for a given period of time to methylate your peptide.

Since you will make the peptide solution yourself, you will know its concentration and volume, so you can calculate how many pmol of arginine are in it, based on the sequence. Suppose you have 0.1 ml of a 10 µM peptide solution containing 2 Arg residues per peptide. The number of pmoles is:

0.0001 L x 10,000,000 pmol/L x 2 = 1000 pmol.

Let's say you decide to let the reaction run for 2 hours.

(1000 pmol Arg /120 min) x [(min-µg enzyme)/0.21 pmol Arg] = 40 µg

Since the enzyme is supplied at 0.47 µg/µl, you need 40 µg x (µl/0.47 µg) = 85 µl.

Thank you for your response!

I think my starting substrate amount of pmol of the peptide with arginine is very high and the final enzyme I will need for the reaction [using your formula] will be a large volume.

I was considering either diluting my substrate or letting my reaction sit for longer.

A follow up question- are there factors I should consider when determining the amount of arginine donor (S-Adenosyl methionine) to add to my reaction?

Based on vendor's testing- 20uM of SAM was used in their assays.

Are there any factors I should consider when deciding how much SAM to add to my reaction?

Thank you

The amount of SAM has to be in excess of the amount of arginine, of course. Also, the product S-adenosylhomocysteine is usually a potent inhibitor of the SAM reactions, so the reaction will slow down as the SAM gets used up. Therefore, you want enough SAM at the start that its concentration won't change very much.

Running the reaction longer will allow you to reduce the amount of enzyme, but only if the enzyme is stable for the whole length of time.

I suggest you try a small-scale reaction first (say, 1/10 scale) to see how well it works before investing a lot of peptide and enzyme.

Thank you Adam B Shapiro ,

It sounds like the best step is to dilute the substrate. If you could read my approach, I would appreciate some feedback.

1. Start with 5ul of a 2.5uM substrate solution > my peptide contains 2 arginines

2. 0.000005L of solution * 5000000 pmol/L (*2 arginine) = 25pmol

3. 25 pmol Arg /120 min) x [(min-µg enzyme)/0.21 pmol/min/ug of Arg = ~1ug

4. Concentration 7.9ug / [0.47ug/ul] = ~2ul of enzyme

The suggested SAM to be added was 20mM [from a 32mM concentrate], which would give me 12.5 ul of SAM + 5ul of substrate + 2ul of enzyme = 20ul reaction that I would let sit for 2 hours. 

I am surprised by the volume of enzyme needed run such a small reaction.

In the future, I am planning to optimize the methylated peptide for an ELISA so I expect some troubleshooting will be needed.

In such a case, would recommend diluting the substrate even more?

In line 4, where it says 7.9 µg, I assume you intended to put ~1 µg.

Earlier, you said the suggested SAM concentration was 20 µM, but in the last entry you said it was 20 mM. Which is it? Is the stock solution 32 mM or 32 µM?

Before deciding whether to dilute the substrate further, consider what concentration you will need for the procedure to follow, and also how much of it you will need.

It's possible that the methylation reaction could be run with less enzyme. You could try some experiments to find out, or perhaps the supplier has some more detailed information available on their web site.

Adam B Shapiro

Apologies for my confusing units!

I was adjusting numbers as I played with my set up and introduced errors.

You are right about being careful with diluting the substrate too much as I eventually plan to do ELISA with my methylated product. My previous ELISA protocols used 0.5mM peptide stock with success. Another option I was considering is to elute the (heavier?) methylated products from the native peptide potentially via spin column. Which would allow me to avoid the substrate dilution and large enzyme volumes.

Thank you.

Corrected approach from previous post:

1. Start with 5ul of a 2.5uM substrate solution > my peptide contains 2 arginines

2. 0.000005L of solution * 2500000 pmol/L (*2 arginine) = 25pmol

3. 25 pmol Arg /120 min) x [(min-µg enzyme)/0.21 pmol/min/ug of Arg = ~1ug

4. Concentration 1ug / [0.47ug/ul] = ~2ul of enzyme

The suggested SAM to be added was indeed 20mM [from a 32mM concentrate], which would give me 12.5 ul of SAM + 5ul of substrate + 2ul of enzyme = 20ul reaction that I would let sit for 2 hours. 

I don't think you will be able to separate the methylated peptide from the unmethylated peptide using a spin column, which is a gel filtration medium. The size difference will be too small. That would be a good way to remove the excess SAM and SAH, however. The yield of peptide might be zero, however, considering the small scale of the reaction we have been discussing.

You could do the peptide separation using a reverse phase HPLC column, probably. By using a volatile mobile phase (water and acetonitrile containing trifluoroacetic acid), you could collect the methylated peptide peak, lyophilize it to dryness, then dissolve it at any concentration you wish. That would also remove the remaining SAM, SAH, and buffer components. But you would have trouble measuring the amount of peptide product at the end considering the tiny amount you started with.

Your best bet would be to run the reaction to completion so there is no unmethylated peptide left, then use it as is for your experiment without purification. You would want to control for the excess SAM in the sample, which is considerable.

Have you asked custom peptide synthesis vendors whether they could chemically synthesize the peptide with methylated arginine?