Defects in Free Fatty Acid Oxidation and Glucose Oxidation Coexist in the Failing Heart
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文摘
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Purpose

Abnormalities of energy substrate utilization exist in the failing heart and contribute to disease progression. These abnormalities are attributed by some to decreased glucose utilization and by others to reduced free fatty acid (FFA) oxidation. In the normal heart, recruitment of the glucose transport proteins GLUT-1 and -4 is a mechanism by which the heart increases glucose transport for metabolism while up- or down-expression of muscle carnitine palmitoyl transferase (mCPT-1) provides directionally concordant regulation of FFA oxidation. In the present study we tested the hypothesis that defects of both GLUT-1 and -4 and mCPT-1 coexist in the failing LV myocardium.

Methods and Materials

Studies were performed in LV tissue homogenate obtained from 6 normal (NL) donor human hearts, 6 explanted failing human hearts with dilated cardiomyopathy (DCM) and 6 failing human hearts with ischemic cardiomyopathy (ICM) and from 6 NL and 7 coronary microembolization-induced heart failure (HF) dogs. Protein levels of the glucose transporters, GLUT-1 and -4 where measured in human and dog hearts and mCPT-1 in dog hearts using specific antibodies and Western blotting and bands quantified in densitometric units (du).

Results

Compared to NL, protein levels of GLUT-1 and -4 were significantly decreased in DCM, ICM and HF dogs and mCPT-1 was significantly decreased in HF dogs ().

Conclusions

Abnormalities of key regulators of FFA oxidation and glucose oxidation coexist in the failing heart regardless of HF etiology and favor energy deprivation. Coexistence of these abnormalities in the failing heart can act to limit the benefits that may be derived from therapies that target.

Protein Expression in LV Myocardium
NL-Human DonorDCMICM
GLUT-1 (du)15.7 ¡À 0.910.6 ¡À 0.3*0.41 ¡À 0.02*
GLUT-4 (du)31.7 ¡À 1.719.5 ¡À 1.8*16.6 ¡À 1.4*
NL-DogHF-Dog
GLUT-1 (du)20.1 ¡À 3.04.7 ¡À 0.5?
GLUT-4 (du)24.5 ¡À 3.116.2 ¡À 0.8?
mCPT-1 (du)1.40 ¡À 0.070.53 ¡À 0.06?

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