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Title: Comparison of cystatin C- and creatinine-based estimation of glomerular filtration rate according to glycaemic status in Type 2 diabetes. Author: Oh SJ, Lee JI, Ha WC, Jeong SH, Yim HW, Son HS, Sohn TS. Journal: Diabet Med; 2012 Jul; 29(7):e121-5. PubMed ID: 22414167. Abstract: AIMS: The influence of hyperglycaemia on the performance of glomerular filtration rate (GFR) estimating equations remains to be determined. We compared the performance of creatinine-based GFR with cystatin C-based GFR in patients with Type 2 diabetes according to glycaemic status. METHODS: In a cross-sectional study of 210 patients with Type 2 diabetes, we staged glycaemic status by HbA(1c) tertiles [HbA(1c) ≤ 75 mmol/mol (9.0%) (n = 70), HbA(1c) 76-95 mmol/mol (9.1-10.8%) (n = 70), HbA(1c) >95 mmol/mol (10.8%) (n = 70)] and measured GFR. Isotopic GFR was measured using renal dynamic imaging with (99m) Tc-diethylene-triamine-penta-acetic acid. Estimated GFR (eGFR) was measured using creatinine-based formulae (Cockcroft-Gault-eGFR, the Modification of Diet in Renal Disease equation-eGFR and the Chronic Kidney Disease Epidemiology Collaboration formula-eGFR) and a cystatin C-based formula (cystatin C-eGFR). RESULTS: The isotopic GFR of all patients was 93.1 ± 34.1 ml min(-1) 1.73 m(-2). All methods for estimating GFR underestimated isotopic GFR [Cockcroft-Gault-eGFR (68.8 ± 38.6 ml min(-1) 1.73 m(-2) ) (P < 0.05), Modification of Diet in Renal Disease-eGFR (74.8 ± 31.3 ml min(-1) 1.73 m(-2) ) (P < 0.05), Chronic Kidney Disease Epidemiology Collaboration-eGFR (72.9 ± 26.6 ml min(-1) 1.73 m(-2)) (P < 0.05) and cystatin C-eGFR (83.5 ± 33.2 ml min(-1) 1.73 m(-2)) (P < 0.05)]. In all patient groups, cystatin C-eGFR was less biased and more accurate than the creatinine-based formulae, especially in the group with HbA(1c) > 95 mmol/mol (10.8%) where there was no difference between cystatin C-eGFR and isotopic GFR. CONCLUSIONS: Performance of cystatin C-eGFR was superior to creatinine-based GFR in patients with Type 2 diabetes with HbA(1c) >95 mmol/mol (10.8%).[Abstract] [Full Text] [Related] [New Search]