关键词: Binding affinity prediction Drug discovery Protein pocket Stacking model

Mesh : Protein Binding Proteins / chemistry Software Drug Development Drug Discovery / methods

来  源:   DOI:10.1016/j.compbiomed.2023.107131

Abstract:
Accurately predicting compound-protein binding affinity is a crucial task in drug discovery. Computational models offer the advantages of short time, low cost and safety compared to traditional drug development. Pocket is the key binding region of the protein, which provides invaluable information for drug repositioning and drug design. In this study, we propose an ensemble learning model, called StackCPA, to predict the compound-protein binding affinity. The model integrates multi-scale features of protein pocket and compound through a transfer learning strategy. The protein pocket is described in a fine-grained way by atomic level, residue level and subdomain level. The proposed model StackCPA is evaluated on three binding affinity benchmark datasets. The experiment results show that StackCPA achieves the best performance on all the three datasets in comparison with other state-of-the-art deep learning models. The ablation study shows that the protein pocket can provide sufficient information for affinity prediction and its multi-scale features enable the model to further improve the prediction performance. In addition, the case study for epidermal growth factor receptor erbB1 (EGFR) indicates that StackCPA could serve as an effective tool for drug repurposing. Source codes and data of StackCPA are available at https://github.com/CSUBioGroup/StackCPA.
摘要:
准确预测化合物-蛋白质结合亲和力是药物发现中的关键任务。计算模型具有时间短的优点,与传统药物开发相比,成本低、安全。口袋是蛋白质的关键结合区,这为药物重新定位和药物设计提供了宝贵的信息。在这项研究中,我们提出了一个集成学习模型,叫做StackCPA,预测化合物-蛋白质结合亲和力。该模型通过迁移学习策略集成了蛋白质口袋和化合物的多尺度特征。蛋白质口袋以原子水平的细粒度描述,残基水平和子域水平。在三个绑定亲和力基准数据集上评估了所提出的模型StackCPA。实验结果表明,与其他最先进的深度学习模型相比,StackCPA在所有三个数据集上都取得了最佳性能。消融研究表明,蛋白质口袋可以为亲和力预测提供足够的信息,其多尺度特征使模型能够进一步提高预测性能。此外,表皮生长因子受体erbB1(EGFR)的案例研究表明,StackCPA可以作为药物再利用的有效工具。StackCPA的源代码和数据可在https://github.com/CSUBioGroup/StackCPA获得。
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