Please find below an example of a calculation. The background on the validation of this method can be found:
“J Clin Endocrinol Metab 84:3666-3672, 1999 – A critical evaluation of simple methods for the estimation of free testosterone in serum”
Concentration Testosterone = FT(free)+ Alb-bound-T + [SHBG]-bound-T
Testosterone = [S] + [SA] + [SP]

· Albumin

[S A] = constant = KA x Conc.Alb = 3.6x104x 43g/l =22.4
[S]6900069000 =(molecular weight alb.)
KA = = 3.6x104
for an average albumin conc. of 4.3 g/dL

or [S A] = 22.43 [S]

[S] + [SA] = (1 + 22.43)[S] = 23.43 [S]     (1)

· SHBG

[P] = free SHBG

[SP] = steroid bound SHBG

K = 109 M

[S] +[P] « [SP] or [S] = [SP] (2)
[P] [K]

[P] + [SP] = [SHBG] or [P] = [SHBG] –[SP]     (3)

·  Bioavailable

[Bio T] = [S] + [SA]
Example :

SHBG : 40 nmol/l = 40 x10-9

Testosterone = 288.4 ng/dl = 10 x 10-9 Mol = [SP] + 23.43 [S]

From (3) [P] = 40 10-9 – 10 10-9 + 23.43 [S] = 30 10-9 + 23.43[S]

From (2) and (1) [S]   =   [SP]   =   10 10-9 – 23.43 [S]   =   10 10-9 – 23.43 [S]
[K] [P] 109 ( 30 10-9 + 23.43[S] ) 30 + 23.4 109[S]

[S]{ 30 +23.43x109 [S]} = 10 10-9 – 23.43[S]

30 [S] + 23.43 x109 [S]2 + 23.43[S] – 10 10-9 = 0

23.43 x109 [S]2 + 53.43[S] - 10 10-9 = 0 second degree equation x =  -b±Öb2-4ac
2a

a = 23.43 x109

b= 53.43

c = - 10 10-9

[S] = -53.43 + Ö 53.432+ 4 x 23.43 109x 10 10-9 = -53.43 + Ö2855+937 = -53.43+61.58
2 x 23.43 x109 48.86 109 48.86 109

[S] = 1.7388 10-10

[S] % = 1.7388-10 x 100 = 1.74 %

FT = 1.7388x 288.5 = 5.02 ng/dl

·  Bioavailable

Bio T = [S] + [SA] = [S] + 22.43 [S]     (for the default albumin concentration of 4.3 g/dL)

Bio T = 5.02 ng/dl x (22.43 + 1) = 118 ng/dL