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Why does aspartimide formation accelerate at room temperature rather than proceeding linearly?

Asked 22 Sept 2024Modified 19 months agoViewed 25k times
23

Setup, so nobody has to ask: aspartimide formation · room temperature.

I want to know whether this is a real physical effect or an artefact of how it is measured.

What prompted the question is an inconsistency between two sources I otherwise trust.

Is the standard explanation correct, and if so, what is the evidence for it?

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SH
askedseven_day_half31k13822 Sept 2024
8Is there a printed date on the vial, and do you know what it was derived from? – Dr_Jonas_Halvorsen 44 days ago
7Voting to keep this open — it is more specific than it first looks. – coring_risk 10 months ago
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4 Answers

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59

Because temperature enters the rate constant through an exponential, so equal steps in temperature multiply the rate instead of adding to it. Arrhenius puts the rate proportional to exp(−Ea/RT); the working approximation is a doubling per 10 K, which takes 5, 15, 25 and 35 °C to multipliers of 1, 2, 4 and 8. The steps in temperature are equal and the steps in rate are not, and that is the whole of the observation. At room temperature the same rule gives about 3.4 times the refrigerated rate, and another 10 K would roughly double it again. A cyclic imide at Asp, eighteen daltons lighter, which then reopens to a mixture including the iso-aspartyl form — same formula as the parent, different molecule, and invisible to a mass-only method. Ea differs by route, so the ranking of routes changes with temperature too — which is why accelerated data extrapolates badly and why nobody should read a 40 °C study as a fast version of a 5 °C one.

Start with the sequence, because which pathways are available depends on which residues are present.

Oxidation targets methionine, cysteine and tryptophan, adding sixteen daltons per oxygen. It is catalysed by trace metals and promoted by dissolved oxygen and by light.

Aggregation is physical: peptides unfold at air-liquid interfaces and associate. Shaking maximises that interface, which is why swirling and shaking produce visibly different outcomes on the same vial.

Sequence determines which pathways apply, so general statements are general.

Swirl, never shake. Aggregation is a handling problem more than a time problem.

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LC
answeredlyoph_cake78k2674 Dec 2024
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39

It helps to be literal here: aggregation is a physical process and is the one most often caused by handling rather than by time.

Hydrolysis cleaves the backbone, most readily at aspartate-proline and aspartate-glycine sequences, and is acid-catalysed. In a dry solid it barely proceeds at all.

More usefully, light exposure matters for tryptophan-containing sequences and for anything with a chromophore. Amber vials and a closed box are free mitigations.

A mass spectrum names the pathway. Plus one, plus sixteen, minus eighteen.

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JF
answeredjuliette_farnese13k3816 Dec 2024
6Aliquoting before the first freeze is the advice I wish I had read two years ago. – lyoph_cake 2 months ago
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28

Answering this needs the physical state, since a dry powder is protected from most of these and a solution is protected from none.

Freeze-thaw cycling drives aggregation through concentration at the ice interface and pH shifts as buffer components crystallise out at different rates. Each cycle costs something.

It helps to be literal here: deamidation converts asparagine or glutamine to the corresponding acid via a succinimide intermediate, adding one dalton. It is base-catalysed, accelerates above neutral pH and is the dominant aqueous pathway for many peptides.

Sequence decides which pathways are even available. Check the residues.

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TG
answeredtandem_gradient61k24812 Nov 2024
22

The short version: water enables most of it, oxygen enables oxidation, surfaces enable adsorption, and agitation enables aggregation.

A mass spectrum resolves most of this: minus eighteen is dehydration or succinimide, plus one is deamidation, plus sixteen is oxidation, and an unchanged mass with a shifted retention time is an isomer.

Apparent loss in a dilute preparation is usually adsorption rather than degradation and is worth ruling out first.

At dilute concentrations, suspect adsorption before you suspect chemistry.

edited 17 Dec 2024 by coldpack_88 — clarified the distinction between purity and content

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C8
answeredcoldpack_8850k3723 Nov 2024

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