Related Experiment Video
Updated: Jul 28, 2026

Measurement of Myocardial Lactate Production for Diagnosis of Coronary Microvascular Spasm
Published on: September 17, 2021
Measurement of extracellular pH, K(+), and lactate in ischemic heart
Sayed A M Marzouk1, Richard P Buck, Larry A Dunlap
1Department of Chemistry, University of North Carolina, Chapel Hill 27599-3290, USA.
Abstract:
Simultaneous and continuous measurements of extracellular pH, potassium (K(+)), and lactate (L(-)) in ischemic rabbit papillary muscle are presented for the first time. Potentiometric pH and K(+) sensors and an amperometric lactate biosensor were used. These miniature electrodes were previously developed and individually tested for this purpose. The pH sensor was based on an iridium oxide layer electrodeposited on a planar platinum electrode fabricated on a flexible substrate. The potentiometric K(+) sensor was based on a polymeric membrane and valinomycin ionophore. The L(-) biosensor was based on lactate oxidase and an organic conducting salt polarized at 0.15V vs Ag/AgCl reference electrode. The utility of this novel analytical system to cardiovascular research was demonstrated by using the system to study the interrelationship of cellular K(+) and lactate loss in ischemic myocardium, and the role of extracellular pH and buffer capacity on this relationship. The results indicated: (i) sequential brief episodes of ischemia produced reproducible trends of L(-), pH, and K(+) changes during the first three episodes, (ii) extracellular L(-) increased with increasing buffer capacity of extracellular compartment, (iii) the patterns of extracellular L(-) and K(+) changes were not related directly, and (iv) L(-) transport and lactic acid diffusion were not the primary cause of extracellular acidosis during ischemia.
![Investigating Cardiac Metabolism in the Isolated Perfused Mouse Heart with Hyperpolarized [1-13C]Pyruvate and 13C/31P NMR Spectroscopy](/_next/image?url=https%3A%2F%2Fcloudfront.jove.com%2FCDNSource%2Fteasers%2F63188.jpg&w=3840&q=50)
