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A two-dimensional speckle-tracking echocardiography for the diagnosis of early myocardial disease in beta-thalassemia major patients
BACKGROUND : Although magnetic resonance imaging T2* is considered the gold standard to assess myocardial iron overload in β-thalassemia patients, its routine use is limited by the high cost and limited availability. Recent data demonstrated that strain imaging by speckle tracking is a sensitive too...
Autores principales: | , , , , |
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Formato: | Online Artículo Texto |
Lenguaje: | English |
Publicado: |
Wolters Kluwer - Medknow
2022
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Materias: | |
Acceso en línea: | https://www.ncbi.nlm.nih.gov/pmc/articles/PMC9802624/ https://www.ncbi.nlm.nih.gov/pubmed/36589651 http://dx.doi.org/10.4103/apc.apc_91_21 |
Sumario: | BACKGROUND : Although magnetic resonance imaging T2* is considered the gold standard to assess myocardial iron overload in β-thalassemia patients, its routine use is limited by the high cost and limited availability. Recent data demonstrated that strain imaging by speckle tracking is a sensitive tool for early assessment of the left ventricular myocardial dysfunction. This study aims to evaluate the clinical utility of two-dimensional (2D) speckle-tracking echocardiography (STE) for the detection of early myocardial disease in beta-thalassemia major (β-TM) patients MATERIALS AND METHODS : 2D STE, magnetic resonance imaging (MRI) heart T2* and MRI liver iron content were done for 30 β-TM patients with no clinical heart disease, compared to 2D STE in 30 healthy age- and sex-matched controls. RESULTS : There was a significant reduction in the longitudinal systolic strain values by STE among β-TM patients compared to controls (P = 0.05). A longitudinal peak systolic strain cutoff values of ≤-19 was able to detect β-TM patients having subclinical cardiac iron overload by MRI T2* (sensitivity = 90%–93.3%, specificity = 83%–100%). Mean serum ferritin in the past 2 years correlated negatively to longitudinal systolic strain values global longitudinal peak systolic strain average (P = 0.05). |
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