These are some articles that provide data on the kinetics of the nonenzymatic hydrolysis of ATP in aqueous solution [Malhotra and Sharma, 1980: (http://www.new.dli.ernet.in/rawdataupload/upload/insa/INSA_1/20005bb0_589.pdf); Couture and Ouellet, 1957: (http://article.pubs.nrc-cnrc.gc.ca/ppv/RPViewDoc?issn=1480-3291&volume=35&issue=11&startPage=1248); Friess, 1952: (http://pubs.acs.org/doi/abs/10.1021/ja01136a016); Seno et al., 1975: (http://joi.jlc.jst.go.jp/JST.Journalarchive/bcsj1926/48.3678?from=Google)]. The article by Friess (1952) provides data on the hydrolysis of tripolyphosphate, which is a three-phosphate "polymer," but it's probably valid to assume that the tripolyphosphoryl- moiety of ATP would undergo nonenzymatic hydrolysis, to ADP and inorganic phosphate (Pi), at a similar rate under the same conditions. And other authors have cited the Friess (1952) article with that assumption in mind. Those articles show that, after 1 hour in solution, at various temperatures, between 91.2 and 99.9 percent of the ATP is still intact (at most, only 8.8 percent of the initial amount/concentration would degrade within 1 hour, based on the data from those articles). This means that storing ATP in solution for days wouldn't be a good idea, but it would be possible to wait for the non-enteric-coated preparations to dissolve in water and then drink the water, to maximize the bioavailability. This can significantly increase the bioavailability of physiological substrates, as shown, in the case of creatine monohydrate, by Deldicque et al. (2008) [Deldicque et al., 2008: (http://www.ncbi.nlm.nih.gov/pubmed/17851680)], discussed here (http://hardcorephysiologyfun.blogspot.com/2009/03/adenosine-and-guanosine-in-animal.html); (http://hardcorephysiologyfun.blogspot.com/2009/04/increase-in-nucleotide-absorption-and.html)]. The rate of entry into solution, in the intestinal luminal fluid, can be a major factor that determines the rate of absorption and, hence, the bioavailability of a high-solubility compound, such as ATP disodium. If a tablet takes 15 minutes to completely dissolve, that has the potential to drastically reduce the bioavailability of ATP disodium. It's partly because of the extremely short half-life of plasma adenosine. If "pre-dissolution" enhanced the bioavailability of creatine, with its much longer half-life, one would expect the rate of dissolution to be even more important for ATP disodium. There could also be greater bioavailability because of a "solvent drag" effect, even though that's mainly been discussed in the context of the enhancement in the rates or extents of absorption of cations, such as magnesium or aluminum, because of the "hydration shells" around those cations. Some ATP is going to be absorbed by passive diffusion, though, and, given the relevance of solvent drag to the bioavailabilities of compounds that are known to be absorbed by passive, paracellular diffusion, it's possible that the phenomenon is relevant to the rate of absorption of ATP [see Kristl and Tukker, 1998: (http://scholar.google.com/scholar?hl=en&q=%22solvent+drag%22+bioavailability)].
I put the calculations below, but it's noteworthy that some researchers have suggested that ATP should be provided in enteric-coated preparations to prevent degradation, prior to absorption, in the stomach or small intestine. Bours et al. (2007a) [Bours et al., 2007a: (http://www.pubmedcentral.nih.gov/picrender.fcgi?artid=1913056&blobtype=pdf)(http://www.ncbi.nlm.nih.gov/pubmed/17578566)] didn't suggest that but nonetheless did use enteric-coated ATP, at a dose of 4 grams/day (4000 mg/day), and suggested that the failure of the ATP to attenuate the toxic effects of indomethacin had occurred because the enteric-coatings on the tablets had dissolved at intestinal sites distal to the site of the toxic effects of indomethacin (Bours et al., 2007a). The authors noted that, in a previous study, ATP administered in solution, directly into the duodenum in humans (at 30 mg/kg bw, twice within a 24-hour period, for a total of 4200 mg/"day," for a 70-kg human) [Bours et al., 2007b: (http://www.ncbi.nlm.nih.gov/pubmed/17301652)], had produced protective effects against the toxic effects of indomethacin (Bours et al., 2007a). The apparent failure of enteric coatings to partially dissolve, in a reliable manner, and allow for the dissolution of tablets [the tablets themselves, beneath the enteric coatings, often fail to dissolve in reasonable amounts of time, too, as laboratory tests have shown, and this is likely to severely limit the bioavailability of something like ATP (and the adenosine derived from it, more precisely)] is not surprising and is consistent with the fact that the pH in the intestinal luminal fluid does not reliably exceed 6.5 until the distal ileum [Fallingborg et al., 1999, discussed here: (http://hardcorephysiologyfun.blogspot.com/2009/04/increase-in-nucleotide-absorption-and.html)].
Malhotra and Sharma (1980) found that the hydrolysis of ATP had followed first-order kinetics, and they found an overall first-order rate constant (I'll call it k1) of 1.52 x 10^(-5) min^(-1) (1.52E-5 min^-1), at 50 degrees C and pH 9.00, and that value for k1 is an overall rate constant that encompasses all of the individual rates of reaction for the individual species that ATP exists as, such as ATP(4-) and HATP(3-), etc. One can write an approximate, general form of the first-order rate equation for the degradation of ATP by keeping the pH constant (more specifically, the concentrations of the hydroxide anion or of protons [the hydrogen ion concentration, H(+) or H3O(+)], respectively, for the base-catalyzed or acid-catalyzed hydrolytic reactions) and, assuming one has calculated the individual rate constants, combining those individual rate constants into a single one (k1). For example, Malhotra and Sharma (1980) simplified the longer form of the rate equation to this:
rate [in mol/(L min) or M/min] = r = k5[ATP(4-)] + k5'[ATP(4-)] ln [OH(-)] = (I'm adding this, as used by Seno et al., 1975, to describe the "pseudo-" first-order kinetics of the nonenzymatic hydrolysis of ATP, meaning that the [H(+)] or [OH(-)] is known and has been incorporated into the pseudo first-order rate constant that I'm calling k1):
= -d[ATP]/dt = k1[ATP], where k1 = something like k5 + k5'ln [OH(-)]
The integration of that gives this (technically, I guess two constants could show up during integration and might have to be combined and incorporated into a final, final rate constant, but I don't need to be precise and am ignoring that):
ln [ATP]t = (-k1)t + ln [ATP]0 (where [ATP]t = the ATP concentration at, time t, in M (mol/L), [ATP]0 = the initial ATP concentration in M (mol/L), t = time elapsed, in min, and k1 = the pseudo first-order rate constant, in min^-1)
So [ATP] = e^((-k1)t + ln [ATP]0), and, at t = 60 minutes, ([ATP]/[ATP]0) x 100 = the percentage of the initial ATP concentration remaining intact after one hour
Couture and Ouellet (1957) collected data showing first-order kinetics, but the way they reported some of the data led me to want to do the calculations using both first-order and zero-order equations. The zero-order rate equation is this:
rate [in mol/(L min) or M/min] = k0 (in this case, the units of the rate constant are M/min, not min^-1, and, given that the rate is constant, are the same as the units of the rate) = -d[ATP]/dt
The integration of that gives this zero-order equation:
[ATP]t = (-k0)t + [ATP]0
So here are the percentages of ATP remaining intact, in solution, after 1 hour:
From Malhotra and Sharma (1980) (I'm using the k value of 1.70E-5 min^-1 (which is similar to the pseudo first-order rate constant), on p. 591, instead of the overall rate constant the authors listed on p. 593, because the authors listed an [ATP]0 value on p. 591:
[ATP]t=60 = e^((-k1)t + ln [ATP]0) = e^((-1.70E-5)(60) + ln (9.218E-3)] = 0.009209 M ATP
([ATP]t=60/[ATP]0) x 100 = ((9.209E-3)/(9.218E-3)) x 100 = 99.902 percent remaining after 1 hour
I'm going to use the kapp value, in Fig. 4 on p. 3679 of Seno et al. (1975), as k1, in this case, and assume it's similar enough to the k0app pseudo first-order rate constant the authors refer to in their equations [the authors mentioned that the kapp and k0app constants were similar under other conditions, and this is essentially the same assumption I mentioned above (that I can ignore any constants that would show up during integration, for the sake of this crude analysis)]. Note that the authors express the constant in these terms:
kapp x 10^7 (in sec^-1) = a number on the graph = 10 at 50 degrees C at pH 4
So kapp = k1 = 10/(1E7) = 1E-6 sec^-1 = (60 sec/1 min)(1E-6/sec) = 6E-5 min^-1
[ATP]t=60 = e^((-k1)t + ln [ATP]0) = e^((-6E-5)(60) + ln (2E-3)] = 1.99281E-3
([ATP]t=60/[ATP]0) x 100 = ((1.99281E-3)/(2E-3)) x 100 = 99.641 percent remaining after 1 hour
The article by Couture and Ouellet (1957) is in French, but I can still glean the data from it (vitesse, in French, means velocity, V, or rate of reaction). Note that the authors report the rate of reaction, not the rate constant, on the dependent axis of Fig. 3, on p. 1251. So:
rate = V = k1[ATP] = -d[ATP]/dt, and I'm going to assume that they're referring to the initial rate. In that case, as shown in Fig. 3, V x 10^9 = 30 m/sec (at pH 8.82), and V = 30/(1E9) = 3E-8 M/sec = 1.8E-6 M/min. So:
Vinitial = 1.8E-6 M/min = k1[ATP]0 = k1(1.175E-3), and k1 = (1.8E-6)/(1.175E-3) = 1.532E-3 min^-1
So:
[ATP]t=60 = e^((-k1)t + ln [ATP]0) = e^((-1.532E-3)(60) + ln (1.175E-3)] = 1.0718E-3 M
([ATP]t=60/[ATP]0) x 100 = ((1.0718E-3)/(1.175E-3)) x 100 = 91.2 percent remaining after 1 hour
I did a zero-order calculation with that reaction rate (the one reported in Fig. 3), and the calculation still showed that about 90 percent of the ATP was remaining after 1 hour. But I think the constant I calculated from the reaction rate, reported in Fig. 3 of Couture and Ouellet (1957), is probably more or less "correct" or valid.
Those data suggest that ATP disodium doesn't need to be prepared in enteric-coated tablets.
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