// May 19th, 2012 // No Comments » // Medical Science

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

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