Identification of Inflamed Atherosclerotic Lesions In Vivo Using PET-CT

Inflammation plays a major pathogenetic role in the development of atherosclerotic plaques and related thromboembolic events. The identification of vulnerable plaques is of the utmost importance, as this may allow the implementation of more effective preventive and therapeutic interventions. Fluorodeoxyglucose positron emission tomography (FDG-PET) has been shown to be useful for tracing inflammation within plaques. However, its relationship to immunohistochemical findings in different territories of the peripheral circulation was not completely elucidated. We aimed to determine whether plaque inflammation could be measured by PET in combination with computer tomography (CT) using FDG and what is the relationship between FDG uptake and immunohistochemical findings in the removed atherosclerotic lesions of the femoral and carotid arteries. The study included 31 patients, 21 patients with high-grade stenosis of the internal carotid artery (ICA) and 10 patients with occlusion of the common femoral artery (CFA), all of whom underwent endarterectomy. Before endarterectomy in all patients, FDG-PET/CT imaging was performed. FDG uptake was measured as the maximum blood—normalized standardized uptake value, known as the target to background ratio (TBR max). TBR max amounted to 1.72?±?0.8, and in patients with ICA, stenosis was not significantly different from patients with CFA occlusion. Immunohistochemical and morphometric analyses of the plaques obtained at endarterectomy showed that the density of T lymphocytes and macrophages (number of cells per square millimeter) was significantly higher in subjects with stenosis of the ICA than in subjects with occlusion of the femoral arteries: lymphocytes, 1.26?±?0.21 vs. 0.77?±?0.29; p?=?0.02 and macrophages, 1.01?±?0.18 vs. 0.69?±?0.23; p?=?0.003. In the whole group of patients, the density of inflammatory cells significantly correlated with FDG uptake represented by PET-TBR max: T lymphocytes, r?=?0.60; p?r?=?0.65; p?