Total amount of pheromones present at any time (and total amount generated in a day mentioned in parantheses) on men who produce the highest amount of pheromones from the study by Nixon A, Mallet AI and Gower DB (see below).
Androstenone = 0.4 (12) mcg
Androstadienone = 3.9 (111) mcg
Beta-Androstadienol = 0.1 (20) mcg
Alpha-Androstenol= 0.5 (47) mcg
Beta-Androstenol = 0.1 (11) mcg
Calculations shown below
From one of my earlier posts:
Simultaneous quantification of five odorous steroi… [J Steroid Biochem. 1988] - PubMed result
One point to note in this publication: “These findings may indicate the existence of a pathway of metabolism in axillary bacteria in which 4,16-androstadien-3-one is reduced to 5 alpha-androst-16-en-3-one and thence to the 3 alpha- and 3 beta-alcohols.”
So effectively it points the possibility of conversion of A1 to androstenone and then to A-nol and B-nol, which means conversion doesn’t just happen from A-nol to -none but may also happen in the opposite direction if I understand correctly.
Now on to the main topic of this post:
Quantity of naturally produced mones
Quantities found (pmol/mg.hair, with approximate 24-h totals in parentheses) were:
Androstenone - 5 alpha-androst-16-en-3-one, 0-15 (0-433)
Androstadienone - 4, 16-androstadien-3-one, 0-143 (0-4103)
Beta-Androstadienol - 5,16-androstadien-3 beta-ol, 0-3.5 (0-728)
Alpha-Androstenol- 5 alpha-androst-16-en-3 alpha-ol, 0-17 (0-1752)
Beta-Androstenol - 5 alpha-androst-16-en-3 beta-ol, 0-4 (0-416)
From the study: "Simultaneous quantification of five odorous steroids (16-androstenes) in the axillary hair of men, " Nixon A, Mallet AI and Gower DB, (Division of Biochemistry, United Medical School, Guy’s Hospital, London, England)
In the current post I will try to convert these numbers to ‘mcg’ units.
- Quantities in the paper are given in pmol/mg.
- 1 pmol = 10^-12 mol (mole).
- 1 mole = the substance’s atomic or molecular mass in grams (For example for C-12 it is 12 grams)
- 1 mole contains a total of N atoms/molecules (N being the Avogadro’s constant = 6.022*10^23).
- As seen from my post on Saturday, September 08, 2007, molecular weight of all these common steroids are very close to 270.
- So for each steroid 1 pmol = 27010^-12 grams = 27010^-9 mg = 270*10^-6 mcg
Therefore for these androstenes
1 pmol = 270*10^-6 mcg
The amounts are given in pmol/mg of hair. So we have to multiply the amounts with total mg of hair on pheromone generating areas of human body (axillary and pubic hair). Lets use a conservative amount for total amount of axillary and pubic hair as 100 mg (QUANTITATIVE MEASUREMENT OF A SECONDARY SEX CHARACTER, AXILLARY HAIR, QUANTITATIVE MEASUREMENT OF A SECONDARY SEX CHARACTER, AXILLARY HAIR - Hamilton - 2006 - Annals of the New York Academy of Sciences - Wiley Online Library says 20.1 mg for axillary hair in under 17, accounting for much more hair after puberty is over we can say it might be approximately 100 mg, but at this point it is a guess only, better estimates are required)
Therefore the total amount of pheromones present at any time (and total amount generated in a day mentioned in parantheses) maybe obtained by multiplying the above quantities given in pmol/mg.hair with (270*10^-6) * (100). This gives,
Androstenone = 0.4 (12) mcg
Androstadienone = 3.9 (111) mcg
Beta-Androstadienol = 0.1 (20) mcg
Alpha-Androstenol= 0.5 (47) mcg
Beta-Androstenol = 0.1 (11) mcg
Not having the amounts of -rone and DHEA-s slightly mars the excitement, but the amounts of given pheromones match reasonably well with what we use.
Finding the ratios of these using the maximum of the 24-h total amount
a-none : a-nol : b-nol : A1 : beta-dienol = 1 : 4.2 : 1 : 9.9 : 1.8
~ 1 : 4 : 1 : 10 : 2
Using these ratios when mixing might help in producing a more natural profile.