Geochimica et Cosmochimica Acta ( IF 5 ) Pub Date : 2022-05-13 , DOI: 10.1016/j.gca.2022.05.006 Kristýna Kantnerová , Shohei Hattori , Sakae Toyoda , Naohiro Yoshida , Lukas Emmenegger , Stefano M. Bernasconi , Joachim Mohn
Multiply substituted isotopic species of nitrous oxide (N2O), referred to as clumped isotopes, represent a promising new tool for distinguishing production pathways of this potent greenhouse gas. This work presents the first determination of enrichment factors of N2O clumped isotopes during bacterial denitrification. Samples of N2O obtained after 1-, 3-, and 7-day incubations of a pure culture of the denitrifier Pseudomonas aureofaciens at 20 °C and 30 °C were analysed by the recently developed quantum cascade laser absorption spectroscopy (QCLAS) method. Enrichment factors εp/s of the cumulative product (p) relative to the substrate (s) were determined using a Rayleigh model for the seven most abundant isotopically substituted molecules (isotopocules) of N2O. Values of the enrichment factors εp/s (with uncertainty expressed as expanded standard uncertainty at the 95% confidence interval) at the two incubation temperatures (20 °C/30 °C) are:
14N15N16O (456): ε456 = (−40.3 ± 2.6)‰/(−35.1 ± 0.7)‰
15N14N16O (546): ε546 = (−38.1 ± 3.4)‰/(−31.2 ± 0.6)‰
14N14N17O (447): ε447 = (21.3 ± 1.2)‰/(24.5 ± 0.5)‰
14N14N18O (448): ε448 = (38.8 ± 1.5)‰/(46.4 ± 1.2)‰
14N15N18O (458): ε458 = (−8.9 ± 2.0)‰/(−11.7 ± 0.6)‰
15N14N18O (548): ε548 = (−3.4 ± 1.1)‰/(−1.8 ± 0.5)‰
15N15N16O (556): ε556 = (−85.9 ± 1.5)‰/(−63.9 ± 1.4)‰
中文翻译:
细菌反硝化形成的一氧化二氮的聚集同位素特征
一氧化二氮 (N 2 O) 的多重取代同位素物种,称为丛集同位素,代表了一种有前途的新工具,可用于区分这种强效温室气体的生产途径。这项工作首次确定了细菌反硝化过程中 N 2 O 聚集同位素的富集因子。通过最近开发的量子级联激光吸收光谱 (QCLAS) 方法分析反硝化菌金黄色假单胞菌纯培养物在 20 °C 和 30 °C 培养 1、3 和 7 天后获得的 N 2 O 样品. 富集因子ε p/s相对于底物 (s) 的累积产物 (p) 使用 Rayleigh 模型确定了七种最丰富的 N 2 O 同位素取代分子(同位素体)。富集因子ε p/s 的值(不确定性表示为两个孵育温度 (20 °C/30 °C) 下 95% 置信区间的扩展标准不确定度为:
14 N 15 N 16 O (456): ε 456 = (-40.3 ± 2.6)‰/(-35.1 ± 0.7)‰
15 N 14 N 16 O (546): ε 546 = (-38.1 ± 3.4)‰/(-31.2 ± 0.6)‰
14 N 14 N 17 O (447): ε 447 = (21.3 ± 1.2)‰/(24.5 ± 0.5)‰
14 N 14 N 18 O (448): ε 448 = (38.8 ± 1.5)‰/(46.4 ± 1.2)‰
14 N 15 N 18 O (458): ε 458 = (-8.9 ± 2.0)‰/(-11.7 ± 0.6)‰
15 N 14 N 18 O (548): ε 548 = (-3.4 ± 1.1)‰/(-1.8 ± 0.5)‰
15 N 15 N 16 O (556): ε 556 = (-85.9 ± 1.5)‰/(-63.9 ± 1.4)‰