Rivlin Michal, Eliav Uzi, Navon Gil
School of Chemistry, Tel Aviv University, Ramat Aviv, Tel Aviv 69978, Israel.
School of Chemistry, Tel Aviv University, Ramat Aviv, Tel Aviv 69978, Israel.
J Magn Reson. 2014 May;242:107-12. doi: 10.1016/j.jmr.2014.02.021. Epub 2014 Mar 4.
Aqueous solutions of formaldehyde, formalin, are commonly used for tissue fixation and preservation. Treatment with formalin is known to shorten the tissue transverse relaxation time T2. Part of this shortening is due to the effect of formalin on the water T2. In the present work we show that the shortening of water T2 is a result of proton exchange between water and the major constituent of aqueous solutions of formaldehyde, methylene glycol. We report the observation of the signal of the hydroxyl protons of methylene glycol at 2ppm to high frequency of the water signal that can be seen at low temperatures and at pH range of 6.0±1.5 and, at conditions where it cannot be observed by the single pulse experiment, it can be detected indirectly through the water signal by the chemical exchange saturation transfer (CEST) experiment. The above finding made it possible to obtain the exchange rate between the hydroxyl protons of the methylene glycol and water in aqueous formaldehyde solutions, either using the dispersion of the spin-lattice relaxation rate in the rotating frame (1/T1ρ) or, at the slow exchange regime, from the line width hydroxyl protons of methylene glycol. The exchange rate was ∼10(4)s(-1) at pH 7.4 and 37°C, the activation energy, 50.2kJ/mol and its pH dependence at 1.1°C was fitted to: k (s(-1))=520+6.5×10(7)[H(+)]+3.0×10(9)[OH(-)].
甲醛水溶液,即福尔马林,常用于组织固定和保存。已知用福尔马林处理会缩短组织横向弛豫时间T2。这种缩短的部分原因是福尔马林对水T2的影响。在本研究中,我们表明水T2的缩短是水与甲醛水溶液的主要成分亚甲基二醇之间质子交换的结果。我们报告了在2ppm处观察到亚甲基二醇羟基质子的信号,该信号位于水信号高频侧,在低温以及pH值为6.0±1.5的范围内可见,并且在单脉冲实验无法观察到的条件下,可通过化学交换饱和转移(CEST)实验通过水信号间接检测到。上述发现使得能够获得甲醛水溶液中亚甲基二醇羟基质子与水之间的交换速率,既可以使用旋转坐标系中的自旋晶格弛豫速率(1/T1ρ)的色散,也可以在慢交换区域,从亚甲基二醇羟基质子的线宽来获得。在pH 7.4和37°C时交换速率约为10(4)s(-1),活化能为50.2kJ/mol,其在1.1°C时对pH的依赖性拟合为:k(s(-1))=520 + 6.5×10(7)[H(+)] + 3.0×10(9)[OH(-)]。