| Literature DB >> 21614163 |
Abstract
This study investigated the effectiveness of Chinese therapeutic food on female reproductive hormones in a double-blind, placebo-controlled clinical trial. Chinese kiwi fruit extract (Hong En No. 1) was provided for Australian peri-menopausal women for one month. Chinese medical assessment and urinary 2-hydroxyestrone (2-OHE) and 16alpha-hydroxyestrone (16alpha-OHE) tests were conducted. Twenty-six urinary samples (pre and post-trial) which met the requirement of testing were analysed, the ratio 2-OHE:16alpha-OHE of pre-trial (1.18 ± 0.34) and post-trial (0.97 ± 0.29) in the control group (n = 6) decreased but showed no significant change, this ratio of pre-trial (1.44 ± 0.16) and post-trial (1.65 ± 0.21) in the treatment group (n = 7) indicated an improvement (P = 0.066), which results in beneficial hormone regulation. The Chinese medicine assessment indicated that the patterns of disharmony mainly include Liver Qi stagnation and Liver-Kidney Yin deficiency patterns. No significant change observed in the control group, significant score reduction of the patterns of disharmony was achieved at post-trial in the treatment group, which indicates an improvement of general health condition.Entities:
Keywords: 2-hydroxyestrone:16α-hydroxyestrone; Chinese kiwi fruit; Chinese medicine patterns of disharmony; Chinese therapeutic food; reproductive hormones
Year: 2011 PMID: 21614163 PMCID: PMC3095195 DOI: 10.4137/IMI.S5782
Source DB: PubMed Journal: Integr Med Insights ISSN: 1177-3936
The effects of Chinese kiwi fruit extract on urinary 2OHE:16α-OHE of peri-menopausal women.*
| Treatment (n = 7) | 47.71 ± 4.68 | 1.44 ± 0.16 | 1.65 ± 0.21 |
| Control (n = 6) | 48.33 ± 6.71 | 1.18 ± 0.34 | 0.97 ± 0.29 |
Notes:
Participants take Chinese kiwi fruit extract (Hong En No. 1) 10 g × 2/days × 4weeks;
P = 0.066, t = 2.244.
Levels of Liver Qi stagnation and Liver-Kidney Yin deficiency at pre and post-trial.
| Treatment (n = 7) | 7.43 ± 1.27 | 2.29 ± 0.49 | 5.14 ± 1.77 | 2.43 ± 1.72 |
| Control (n = 6) | 7.00 ± 1.10 | 7.20 ± 0.98 | 5.17 ± 1.94 | 5.17 ± 1.94 |
Notes:
P = 0.00, t = 12.73;
P = 0.00, t = 7.55.