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Chapter 2: Problem 10

The \(\mathrm{LD}_{50}\) for a potential antiobesity compound was found to be \(10 \mathrm{mg} / \mathrm{kg}\) and the \(\mathrm{ED}_{50}\) was \(2 \mathrm{mg} / \mathrm{kg}\). Is this an important drug candidate? Why or why not?

Short Answer

Expert verified
Given the LD50 to ED50 ratio of 5, the drug can be deemed safe and could be considered as a potential candidate.

Step by step solution

01

Understand the Concepts

LD50: It is the dose required to kill 50% of the population. Here, the given LD50 is \(10 \mathrm{mg} / \mathrm{kg}\). ED50: It is the dose required for the drug to be effective in 50% of the population. Here, the given ED50 is \(2 \mathrm{mg} / \mathrm{kg}\). What this means is that, 2mg/kg is enough to have intended effect in the population while 10mg/kg is lethal to the same population. Mapping this understanding to the given problem is key.
02

Identify LD50 to ED50 Ratio

The next step is to compare the LD50 value with the ED50 value. We could establish a ratio - divide the LD50 by the ED50. This is \(10 \mathrm{mg} / \mathrm{kg}\) divided by \(2 \mathrm{mg} /\mathrm{kg}\), giving a result of 5.
03

Evaluate the Drug's Safety

If the resulting ratio is large, the drug is deemed safer because the lethal dose is much higher than the effective dose, which means it would take a lot more of the drug to risk being lethal. In this case, the ratio is 5. That is, the lethal dose is five times higher than the effective dose. This is a relatively high ratio, indicating that the drug could be considered relatively safe.

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