I was just playing around with some numbers one day after a lecture for school and came up with some formula for calculating the additive emetic level of more than one antineoplastic drug. I thought I’d share it here in case anyone thought it was interesting (and, well, there’s little else to share). Maybe someone’s doing a Google search and would think this is cool like I did… (I didn’t know what else to do with this since I never got a chance to use it for the exam :-/ Garrrr…)

Antineoplastics are categorized according to their emetogenic potential, Frequency of emesis
▫ Level 1 (<10%), Busulfan, Vincristine, Chlorambucil
▫ Level 2 (10-30%), Gemcitabine, Paclitaxel, Asparaginase
▫ Level 3 (30-60%), Cyclophosphamide, Methotrexate
▫ Level 4 (60-90%), Carmustine, Cisplatin, Irinotecan
▫ Level 5 (>90%), Mechlorethamine, Dacarbazine
Combining antineoplastics may increase emetogenic risk:
For example:
(1) 2 + 2 + 2 = 3
(2) 2 + 2 + 3 = 4
(3) 3 + 3 + 3 = 5
If you have 3 drugs that give about a Level 2 Emesis (10-30%), the aggregate that the patient would experience would be 30-60% (More Emetogenic)

If you’re given a list of drugs and need to find the additive emesis level for that list based on the level of each drug:
Let a = Minimum Range Number for a given drug’s Level number
Let b = Maximum Range Number for a given drug’s Level number
Let m = The percentage of emesis for a given drug (1 singular drug).
Let n = The number of drugs in the list given (if 1 drug, n=1; if 2 drugs, n=2, etc.)
Let p = The sum of the percentages for a number of drugs (many drugs)

Formula #1: m = (b/2)
— For a given drug “n” Formula #2: p = sum(m) + 1
— Letter j represents Then, answer = the level of p.

2 + 2 + 2 = 3 – There are three Level 2 drugs given.
Therefore, n=3.
– Level 2’s Range is 10-30%.
Therefore, a = 10, b = 30
– m1 = (30/2) = 15%
– m2 = (30/2) = 15%
– m3 = (30/2) = 15%
– p = sum(15% + 15% + 15%) + 1 = 46%
The answer is “3” since 46% is in the Level 3 range

2 + 2 + 3 = 4 – There are two Level 2 drugs and one Level 3 drug given.
Therefore, n = 3.
– Level 2’s Range is 10-30%.
– Level 3’s Range is 30-60%.
– m1 = (30/2) = 15%
– m2 = (30/2) = 15%
– m3 = (60/2) = 30%
– p = sum(15% + 15% + 30%) + 1 = 61%
The answer is “4” since 61% is in the Level 4 range.

3 + 3 + 3 = 5 – There are three Level 3 drugs and one Level 3 drug given.
Therefore, n = 3. – Level 3’s Range is 30-60%.
– m1 = (60/2) = 30%
– m2 = (60/2) = 30%
– m3 = (60/2) = 30%
– p = sum(30% + 30% + 30%) + 1 = 91%
The answer is “5” since 91% is in the Level 5 range.

Hope that amuses someone out there. -ID