Figure - PMC

Skip to main content

An official website of the United States government

Here's how you know

Here's how you know

Official websites use .gov
A .gov website belongs to an official government organization in the United States.

Secure .gov websites use HTTPS
A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites.

View full-text article in PMC

. 2019 Dec 4;9:18293. doi: 10.1038/s41598-019-54591-3

Search in PMC
Search in PubMed
View in NLM Catalog
Add to search

© The Author(s) 2019

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons license, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this license, visit http://creativecommons.org/licenses/by/4.0/.

PMC Copyright notice

Multivariate statistical analysis distinguished the phenolic compounds on the antibacterial activity against Staphylococcus aureus. (A) OPLS-DA and (B) PLS-DA analysis during the rainfall and drought periods. (C) VIP analysis (Variable Importance in Projection) extracted from the PLS-DA plot identified principal phenolic compounds (guttiferones) and classes of flavonoids (flavanones) responsible by anti-staphylococcal activity during the seasonal study. The OPLS-DA model was validated by CV-ANOVA, p = 0.000004 (p < 0.05), F = 12.47. The PLS-DA model was validated by CV-ANOVA, p = 0.000003 (p < 0.05), F = 12.90.