| Literature DB >> 23554712 |
Ali Ahadi1, Alireza Partoazar, Mohammad-Hassan Abedi-Khorasgani, Seyed Vahid Shetab-Boushehri.
Abstract
Liquid-liquid extraction-thin layer chromatography (LLE-TLC) has been a common and routine combined method for detection of drugs in biological materials. Solid-phase extraction (SPE) is gradually replacing the traditional LLE method. High performance thin layer chromatography (HPTLC) has several advantages over TLC. The present work studied the higher efficiency of a new SPE-HPTLC method over that of a routine LLE-TLC method, in extraction and detection of urinary morphine. Fifty-eight urine samples, primarily identified as morphine-positive samples by a strip test, were re-screened by LLE-TLC and SPE-HPTLC. The results of LLE-TLC and SPE-HPTLC were then compared with each other. The results showed that the SPE-HPTLC detected 74% of total samples as morphine-positive samples whereas the LLE-TLC detected 48% of the same samples. We further discussed the effect of codeine abuse on TLC analysis of urinary morphine. Regarding the importance of morphine detection in urine, the present combined SPE-HPTLC method is suggested as a replacement method for detection of urinary morphine by many reference laboratories.Entities:
Keywords: high-performance thin layer chromatography; liquid-liquid extraction; morphine detection; solid-phase extraction; thin-layer chromatography
Year: 2011 PMID: 23554712 PMCID: PMC3596733 DOI: 10.1016/S1674-8301(11)60048-1
Source DB: PubMed Journal: J Biomed Res ISSN: 1674-8301
Comparison between the results obtained by LLE-TLC and SPE-HPTLC in 58 urine samples primarily identified as morphine-positive samples by ACON® MOP One Step Opiate Test Strip.
| LLE-TLC ( | ||||
| Positive | Negative | Total | ||
| SPE-HPTLC | Positive | 26 | 17 | 43 |
| Negative | 2 | 13 | 15 | |
| Total | 28 | 30 | 58 | |
LLE-TLC: Liquid Liquid Extraction-Thin Layer Chromatography, SPE-HPTLC: Solid Phase Extraction-High Performance Thin Layer Chromatography. McNemar's χ2= 9.8, κ = 0.355, P = 0.001