After assessing an LD50 of a toxin, now need to evaluate the effect of an antidote against the toxin, can doubling the value of the LD50 be equal to the LD100 of the toxin so as to ascertain the efficacy of the antidote?
Unfortunately, logic does not give the correct mathematical result. The lethality curves are never completely linear, and usually a curve, which needs to be determined by study. There may be curves available for at least the common toxins, if one is willing to search. So the answer here, logically maybe, but practically, no. Thank you.
Yes, it can be if you don't take LD50 to be equal to median survival dose 50. Kindly refer to publication by Saganuwan entitled :"Revised arithmetical method of Reed and Muench for deternination of a relatively ideal LD5 and other similar articles published by the same author.
Unfortunately, logic does not give the correct mathematical result. The lethality curves are never completely linear, and usually a curve, which needs to be determined by study. There may be curves available for at least the common toxins, if one is willing to search. So the answer here, logically maybe, but practically, no. Thank you. " Gary Joseph Ordog ", This is what I want to recommend for you dear.
Why do you need an LD100 to test an antidote? Any known and replicable value such as LD50 or even LD 10 can give you statistically assessable response to determine antidote efficacy. Based on modern animal research regulations, an LD100 test would not be accepted as it unnecessarily kills too many animals. A well controlled experiment possibly using anesthetized animals with physiological monitoring can provide a lot of information with very few animals used depending on the nature of your toxin-antidote combination.
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