Postoperative coronal plane imbalance and model construction for adolescent idiopathic scoliosis
10.3969/j.issn.2095-4344.2014.09.009
- VernacularTitle:青少年特发性脊柱侧凸植入物内固定后冠状面失平衡及模型构建
- Author:
Shihao LI
;
Qiang DENG
;
Chuanhui XUN
;
Weibin SHENG
- Publication Type:Journal Article
- Keywords:
internal fixator;
scoliosis;
regression analysis;
linear model
- From:
Chinese Journal of Tissue Engineering Research
2014;(9):1362-1367
- CountryChina
- Language:Chinese
-
Abstract:
BACKGROUND:Adolescent idiopathic scoliosis is a common disease that can affect physical appearance of adolescents in the clinic at present. However, there are lacks of studies on coronal plane imbalance after fixation using Logistic regression equation at present.
OBJECTIVE:To investigate the reasons for coronal plane imbalance after fixation in patients with Lenke type II adolescent idiopathic scoliosis.
METHODS:141 cases of Lenke type II adolescent idiopathic scoliosis admitted by Department of Spinal Surgery of the First Affiliated Hospital of Xinjiang Medical University in China from January 2001 to November 2012 were chosen as subjects. Multivariate single factor and multiple-factor Logistic regression were performed. Risk factors for the coronal plane imbalance after fixation in adolescent idiopathic scoliosis patients were screened, and predictive models were established.
RESULTS AND CONCLUSION:Coronal plane imbalance occurred in 30 of the 141 patients, accounting for 21.28%. For Lenke type II adolescent idiopathic scoliosis patients, preoperative apical vertebral Nash-More rotation level 3-4, Risser grade 4-5, major curve correction rate/flexibility>1, lower thoracic Cobb angle>70° were vulnerable to postoperative coronal plane imbalance. Multivariate logistic regression showed that vertebral rotation, Risser grade, major curve correction rate/flexibility, lower thoracic Cobb angle were independent risk factors for postoperative coronal plane imbalance in Lenke type II adolescent idiopathic scoliosis patients. The predictive model was Y=1/[1+exp(-1.182X 1+1.228X 2+1.671X 3-0.71X 4+0.407)].
