Posts Tagged ‘medical science’

An Illustrative Parallel between the Bible & DNA

// August 16th, 2013 // No Comments » // Blog

The cell’s entire play has one script — DNA. The wisdom behind the construction of biological systems is profound. The cell treats DNA like the Bible — with great reverence. Most proteins are synthesized off of an mRNA copy of the DNA rather than directly from the DNA itself, and (like the Bible’s central role in determining the life of a follower of Christ) the cell refers studiously to the DNA for direction of literally every cellular process. There’s a gene that controls literally everything, just like scripture is sufficienct for all of life and godliness for the Christian. Just a thought.

Could this be a cure for Malaria?

// August 9th, 2013 // No Comments » // Blog

The U.S. military and the medical community have just announced that they’re sitting on a potential cure to Malaria. 3.3 billion people live in areas where they are at a risk of getting Malaria. There were 219 million cases in 2010 (about 660,000 people around the world lost their lives to Malaria). So, this is a huge dent into the mortality rate of Malaria to have a 100% cure. For one year as an undergraduate I did undergraduate research in a lab working for MMV (Medical Malaria Ventures) where we did extractions on thousands of specimens to send them off to a microbiological lab for Kirby-Bauer Drug Testing to find potential hits. It was exciting to think that one of them could be a potential cure, in my hands. Brute-force drug discovery is very hard, but needed. I’m glad they found a cure That’s very good. But what are they going to do about the world’s leading cause of death — abortion?

This is what CNN has to say about the Malaria Vaccine: http://www.cnn.com/2013/08/08/health/malaria-vaccine/index.html

 

Strange find: Physiologic Necrosis

// August 3rd, 2013 // No Comments » // Blog, Medical Science

“…under specific conditions such as caspase intervention, necrosis has been proposed to be regulated in a well-orchestrated way as a backup mechanism of apoptosis. The term programmed necrosis has been coined to describe such an alternative cell death.” http://www.ncbi.nlm.nih.gov/pubmed/20390367

Math Meets Medical Science – Additive Emetogenic Potential

// 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