Retinal thickness decreases with age: an OCT study
Retinal thickness decreases with age: an OCT study
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DOI：
10.1136/bjo.87.7.899
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发表时间：
2003-07-01
影响因子：
4.1
通讯作者：
Funk, J
中科院分区：
文献类型：
--
作者：
Alamouti, B;Funk, J
背景/目的：在三维视盘断层扫描中，需要参考平面来计算视盘边缘或杯值。参考平面的位置通常取决于颞盘边缘的视网膜厚度。最初，人们认为颞盘边缘的视网膜厚度与年龄无关。本研究的目的是检查这一假设，使用光学相干断层扫描，并另外确定在这方面的OCT测量的再现性。方法：100眼100名健康志愿者被纳入本研究。对每只眼睛进行三次OCT扫描。扫描垂直对齐，并放置在视盘的颞侧边缘。对于每只眼睛，计算整个视网膜的厚度以及视网膜神经纤维层的厚度及其变异系数。此后视网膜厚度和神经纤维层厚度与年龄。结果：平均视网膜厚度为249（SD 22）妈妈。平均神经纤维层厚度为109（22）μ m。平均变异系数为3.5%（视网膜总厚度）和8.0%（神经纤维层厚度）。视网膜总厚度和神经纤维层厚度均与年龄显著相关（视网膜总厚度：y = 269.5 - 0.53 x x; R-2 = 0.133; p = 0.0002;神经纤维层厚度：y = 126.8 - 0.44 x x; R-2 = 0.094; p
Background/aim: In three dimensional optic disc tomography a reference plane is required to calculate optic disc rim or cup values. The position of the reference plane often depends on the retinal thickness at the temporal disc margin. Originally it was assumed that the retinal thickness at the temporal disc margin is independent of age. The aim of the study was to check this hypothesis using optical coherence tomography, and additionally to determine the reproducibility of OCT measurements in this area.Methods: 100 eyes of 100 healthy volunteers were included in this study. Three OCT scans were performed on each eye. The scans were aligned vertically and placed at the temporal edge of the optic disc. For each eye, the thickness of the whole retina as well as the thickness of the retinal nerve fibre layer were calculated together with their coefficients of variation. Thereafter retinal thickness and nerve fibre layer thickness were correlated with age.Results: The mean retinal thickness was 249 (SD 22) mum. The mean nerve fibre layer thickness was 109 (22) mum. The mean coefficients of variation were 3.5% (total retinal thickness) and 8.0% (nerve fibre layer thickness). Both the total retinal thickness and the nerve fibre layer thickness were significantly correlated with age (total retina: y = 269.5 - 0.53 x x; R-2 = 0.133; p = 0.0002, nerve fibre layer: y = 126.8 - 0.44 x x; R-2 = 0.094; p