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Measuring Stolons and Rhizomes of Turfgrasses Using a Digital Image Analysis System
Published on: February 19, 2019
Quality evaluation of Kaempferia parviflora rhizome with reference to 5,7-dimethoxyflavone
Yamon Pitakpawasutthi1, Chanida Palanuvej1, Nijsiri Ruangrungsi1,2
1Public Health Sciences Program, College of Public Health Sciences, Chulalongkorn University, Bangkok 10330, Thailand.
Abstract:
Kaempferia parviflora Wall. ex Baker is a medicinal plant found in the upper Northeastern regions of Thailand, which belongs to Zingiberaceae family. The present study aims to investigate the standardization parameters, to analyze chemical constituents of volatile oil by gas chromatography-mass spectrometry, and to determine the content of 5,7-dimethoxyflavone in K. parviflora rhizomes by thin-layer chromatography (TLC)-densitometry compared to TLC image analysis. K. parviflora rhizomes from 15 different sources throughout Thailand were investigated for morphological and pharmacognostic parameters. 5,7-Dimethoxyflavone contents were determined by TLC-densitometry with winCATS software and TLC image analysis with ImageJ software. The mobile phase for TLC development consisted of toluene: chloroform: Acetone: formic acid (5: 4: 1: 0.2). For the Results, the pharmacognostic parameters of K. parviflora rhizome were demonstrated. The loss on drying, total ash, acid-insoluble ash, water content, volatile oil content, ethanol, and water-soluble extractive values were found to be 8.979 ± 0.041, 5.127 ± 0.060, 2.174 ± 0.092, 9.291 ± 0.458, 0.028 ± 0.003, 5.138 ± 0.092, and 8.254 ± 0.191 g/100 g of dry weight, respectively. K. parviflora volatile oil showed the major components of α-copaene, dauca-5, 8-diene, camphene, β-pinene, borneol, and linalool. The 5,7-dimethoxyflavone content of K. parviflora rhizomes determined by TLC-densitometry and TLC image analysis were found to be 2.15 ± 0.64 and 1.96 ± 0.51 g/100 g of dry rhizomes, respectively. The 5,7-dimethoxyflavone contents of both methods were not significantly different (P > 0.05) using paired t-test.
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