Biology Stack Exchange is a question and answer site for biology researchers, academics, and students. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a dose-response experiment testing a new cancer drug. the darker line represents cancer cells. what range of dose should be used? I think it's 2-4 because this affects cancer cells only. is this correct?

enter image description here

share|improve this question
Assuming that that's % cells rather than # then the ideal dose would be the minimum dose that killed 100% of cancer cells - minimum because you want to stay far away from the healthy cells line. This would be a pretty amazing drug however, in real life those lines are likely to overlap long before 100% cancerous cell death, hence the problems of dosing chemotherapy. – Rory M Nov 11 '12 at 23:05

In the most basic sense you want to kill the most cancerous cells whilst minimizing the regular somatic cell death. Almost all cancer medications affect regular cells, too - though the better ones do so at a minimum whilst being effective. In reality, it's also nearly impossible to kill all of the cancerous cells. The goal is to bring them below detectable levels, which can allow the body to finish the job. Leaving significant amounts of cancerous cells alive won't do the patient any good - they'll just continue to proliferate and the patient will be back for more operations or treatments soon.

So, with the goal of minimizing benign cell cost and completely eradicating the cancerous line, on your crude chart that falls at about "4".

share|improve this answer
so what would the range be? 0-4 or 2-4? – Kirby Nov 11 '12 at 23:14
Probably 3-4. A dose of "2" isn't killing significant amounts of cancerous cells. Though everything is a bit vague about the question you're asking. Normally doses are determined by mg/kg depending on the utilization and persistence of the drug in the body. – MCM Nov 11 '12 at 23:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.