Predicting RNA secondary structures including pseudoknots by covariance with stacking and minimum free energy.
- Author:
Jinwei YANG
1
;
Zhigang LUO
;
Xiaoyong FANG
;
Jinhua WANG
;
Kecheng TANG
Author Information
1. College of Computer Science, National University of Defense Technology, Changsha 410073, China.
- Publication Type:Journal Article
- MeSH:
Algorithms;
Base Pairing;
Base Sequence;
Computational Biology;
methods;
Molecular Sequence Data;
Nucleic Acid Conformation;
RNA;
chemistry;
genetics;
Sequence Analysis, RNA
- From:
Chinese Journal of Biotechnology
2008;24(4):659-664
- CountryChina
- Language:Chinese
-
Abstract:
Prediction of RNA secondary structures including pseudoknots is a difficult topic in RNA field. Current predicting methods usually have relatively low accuracy and high complexity. Considering that the stacking of adjacent base pairs is a common feature of RNA secondary structure, here we present a method for predicting pseudoknots based on covariance with stacking and minimum free energy. A new score scheme, which combined stacked covariance with free energy, was used to assess the evaluation of base pair in our method. Based on this score scheme, we utilized an iterative procedure to compute the optimized RNA secondary structure with minimum score approximately. In each interaction, helix of high covariance and low free energy was selected until the sequences didn't form helix, so two crossing helixes which were selected from different iterations could form a pseudoknot. We test our method on data sets of ClustalW alignments and structural alignments downloaded from RNA databases. Experimental results show that our method can correctly predict the major portion of pseudoknots. Our method has both higher average sensitivity and specificity than the reference algorithms, and performs much better for structural alignments than for ClustalW alignments. Finally, we discuss the influence on the performance by the factor of covariance weight, and conclude that the best performance is achieved when lambda1 : lambda2 = 5 : 1.
