Kotikova A I, Blinova E A, Akleyev A V
Urals Research Center for Radiation Medicine, Federal Medical-Biological Agency of the Russian Federation, Chelyabinsk, Russia.
Chelyabinsk State University, Chelyabinsk, Russia.
Dokl Biochem Biophys. 2025 Apr;521(1):261-266. doi: 10.1134/S1607672925700048. Epub 2025 Apr 12.
To study the repertoire of the T-cell receptor in persons chronically exposed to radiation in the long-term period.
The study involved 48 people, who were divided into two groups: a group of exposed persons (31 individuals with the average accumulated dose to red bone marrow (RBM) of 981 ± 130 mGy) and a comparison group (17 individuals with the average accumulated dose to RBM of 25.3 ± 5.91 mGy). The study groups did not differ significantly in age, gender and ethnicity. The repertoire of Vβ-segments of the T-cell receptor of the peripheral blood T-lymphocytes of exposed persons was analyzed by flow cytometry method. 24 Vβ-segments of the T-cell receptor were studied. Statistical processing of the obtained data was carried out using the Wilcoxon signed-rank test, and a direct description of Vβ-segment repertoire of the T-cell receptor was performed using the Lorenz curve and the Gini-TCR index.
The study revealed a statistically significant increase in the number of Vβ3 and Vβ5.2 T-cell receptor segments in exposed individuals relative to the comparison group (p = 0.03 and p = 0.003, respectively). It was also shown that the distribution of the Vβ-segments of the T-cell receptor is uneven in both study groups. However, there was no significant difference between the repertoires of the T-cell receptor of the studied groups in the Gini-TCR index (p = 0.14).
研究长期受到慢性辐射的人群中T细胞受体库情况。
该研究纳入48人,分为两组:暴露组(31人,红骨髓平均累积剂量为981±130毫戈瑞)和对照组(17人,红骨髓平均累积剂量为25.3±5.91毫戈瑞)。研究组在年龄、性别和种族方面无显著差异。采用流式细胞术分析暴露组外周血T淋巴细胞T细胞受体Vβ片段库。研究了24个T细胞受体Vβ片段。使用Wilcoxon符号秩检验对所得数据进行统计处理,并使用洛伦兹曲线和基尼-TCR指数对T细胞受体Vβ片段库进行直接描述。
研究显示,与对照组相比,暴露个体中T细胞受体Vβ3和Vβ5.2片段数量有统计学意义的增加(分别为p = 0.03和p = 0.003)。还表明,两个研究组中T细胞受体Vβ片段的分布均不均匀。然而,研究组T细胞受体库在基尼-TCR指数方面无显著差异(p = 0.14)。