Meta-analysis of the effect of structured exercise training on cardiorespiratory fitness in Type 2 diabetes mellitus

Meta-analysis of the effect of structured exercise training on cardiorespiratory fitness in Type 2 diabetes mellitus
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DOI:
10.1007/s00125-003-1160-2
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发表时间:
2003-08-01
期刊:
影响因子:
8.2
通讯作者:
Sigal, RJ
Sigal, RJ
中科院分区:
医学1区
文献类型:
--
作者:
Boulé, NG;Kenny, GP;Sigal, RJ

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目的/假设。低心肺适应性是糖尿病患者死亡率的一个强有力的独立预测因素。几项研究已经检查了运动对2型糖尿病患者心肺功能的影响。然而,这些研究的样本量相对较小,结果差异很大。因此,本研究的目的是系统地回顾和量化运动对2型糖尿病患者心肺功能的影响。我们检索了截至2002年3月的MEDLINE、EMBASE和其他四个数据库,以寻找随机对照试验,评价8周或更长时间的结构性有氧运动干预对2型糖尿病成人心肺适能的影响。呼吸适应性定义为最大运动试验中的最大摄氧量(VO 2 max)。7项研究,提供了比较运动组和对照组的9项随机试验的数据(总n=266),符合纳入标准。平均运动特征如下:每周3.4次,每次49分钟,持续20周。运动强度范围为VO 2 max的50%至75%。运动组的最大摄氧量增加了11.8%,对照组减少了1.0%(干预后标准化平均差异=0.53,p
Aims/hypothesis. Low cardiorespiratory fitness is a powerful and independent predictor of mortality in people with diabetes. Several studies have examined the effects of exercise on cardiorespiratory fitness in Type 2 diabetic individuals. However, these studies had relatively small sample sizes and highly variable results. Therefore the aim of this study was to systematically review and quantify the effects of exercise on cardiorespiratory fitness in Type 2 diabetic individuals.Methods. MEDLINE, EMBASE, and four other databases were searched up to March 2002 for randomized, controlled trials evaluating effects of structured aerobic exercise interventions of 8 weeks or more on cardiorespiratory fitness in adults with Type 2 diabetes. Cardiorespiratory fitness was defined as maximal oxygen uptake (VO2max) during a maximal exercise test.Results. Seven studies, presenting data for nine randomized trials comparing exercise and control groups (overall n=266), met the inclusion criteria. Mean exercise characteristics were as follows: 3.4 sessions per week, 49 min per session for 20 weeks. Exercise intensity ranged from 50% to 75% of VO2max. There was an 11.8% increase in VO2max in the exercise group and a 1.0% decrease in the control group (post intervention standardized mean difference =0.53, p