Why would a patient with cyanide poisoning have elavated serum lactate?

Answer 1

Cyanide prevents the use of oxygen at the cytochrome oxidase at the mitochondria in the cell. This results in very rapid change to anaerobic metabolism by the cell.

Cyanide prevents the use of oxygen at a critical site (Aa3) on the electron transport chain. The end result is excess lactate in the muscle cells. Since muscle cells cannot utilize lactate it is released into the serum where it can be metabolized in the liver back to glucose. The very high levels of serum lactate reflect the very highly toxic nature of cyanide. It is bound tightly and irreversibly to the Aa3 rapidly causing anaerobic metabolism. While other poisons may attach to Aa3 (Carbon Monoxide) they are not as tightly bound, and thus not as toxic.

A more detailed answer is as follows.

Oxygen is not allowed the attach to the Aa3 (cytochrome oxidase) site on the electron transport chain. Electrons released by the upper sites on the chain are unable to be utilized, and the chain backs up with free electrons. NADH and FADH2 cannot oxidize into NAD+ and FAD, thus the whole electron chain is inactivated. Because NAD+ and NADH become unavailable the Krebs cycle in inactivated. The glycolysis pathway continues to function, but Pyruvate acid must be changed to Lactic acid to generate enough NAD+ to keep the gylcolysis pathway functioning. The end result is only 2 ATP available instead of the normal 36. Cellular function rapidly fails and lactic acid very rapidly increases.

Very high ( greater than 10 mmol/L) levels of Lactate indicate almost a complete failure of the metabolic pathway. Cyanide is the one poison that can render this system inactive so rapidly