Protective index: Difference between revisions

Revision as of 14:00, 28 January 2024

Revision as of 14:07, 28 January 2024

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dose of a toxicant must be multiplied to produce a

defined level of toxicity in the presence of a nontoxic

dose of another chemical<ref name="Bhat et al. 2007">{{cite journal |last=Bhat |first=A. Shakoor |last2=Ahangar |first2=Azad A. |date=2007-02-03 |title=Methods for Detecting Chemical–Chemical Interaction in Toxicology |url=http://www.tandfonline.com/doi/full/10.1080/15376510601177654 |journal=Toxicology Mechanisms and Methods |language=en |volume=17 |issue=8 |pages=441–450 |doi=10.1080/15376510601177654 |issn=1537-6516}}</ref>

dose of another chemical^{^{Bhat et al., 2007}}.

<math>\mbox{Protective index} = \frac{\mathrm{LD}_{50}\text{of A with B}}{\mathrm{LD}_{50}\text{of A alone}}</math>

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Since the protective index is higher for the simultaneous use of nitrite and thiosulfate, the

two chemicals constitute the antidote against cyanide

intoxication.<ref name="Bhat et al. 2007" />

Bhat, A Shakoor and Ahanger, Azad A, 2007. Toxicology Mechanisms and Methods, 17:441–450, 2007

{{Pharmacology}}