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Topic: Molality Vs Molarity when looking at protein folding?  (Read 2258 times)

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Offline Denverboard

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Molality Vs Molarity when looking at protein folding?
« on: July 14, 2015, 01:49:14 PM »
I am studying the folding pathway of a protein and using several denaturants to do so. I've been keeping my denaturants at fixed molalities because a lot more thermodynamic formulas use molality instead of the classical molarity. I have also been using a phosphate buffer at a set molarity. I'm looking at the effect my buffer concentration has on the amount of denaturant needed to unfold the protein. I've noticed that some of the denaturants change the volume of the final solution and therefore change the molarity of my buffer. However, I have calculated that my buffer molality is staying constant throughout the experiment. So, I am wondering if anyone here has any thoughts on whether it is better to record my buffer by its molarity or molality while observing the effect it has on the amount of denaturant needed to unfold the protein. Once again the denaturant is being set at varying molalities.

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Offline MrTeo

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Re: Molality Vs Molarity when looking at protein folding?
« Reply #1 on: July 15, 2015, 04:24:29 PM »
I find it hard to believe that while the molarity of your buffer in your sample changes as you add denaturant the molality doesn't as they're easily connected using their definitions so if one changes the other one does too (as you increase the volume of the solution the mass of the solvent will also increase).
I should probably know something more about your experiment to help, but if the denaturing agent is something like urea or guanidinium chloride I don't see any other reason that its effect as a salt why the buffer concentration should change your protein's behaviour. And if that is the case you can simply add a different salt to the solution to see if it destabilizes your protein or, maybe, just helps it resist the increasing denaturant concentration.

The reason why molality is often used in thermodynamics, anyway, is because it is volume (and thus temperature) independent and, in the range where most of the usual thermodynamic relationships for solutions are used (e.g. Raoult's law, Henry's law,... so when one of the molar fractions of our two-component mixture is ≪1), its value is directly proportional to the molar fraction (which is what you use to express and derive pretty much everything in solution thermodynamics).
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