Supplemental Materials and Methods. Derivations of Calculated Free and Bioavailable25(OH)D* from Powe et al. J Bone Miner Res 2011; 26:1609-1616 (ref. 21)
*For the purpose of these calculations, [Total D] is used to designate total-25(OH)D and [Bio D]
to designate bioavailable-25(OH)D.
DEFINITIONS
[Total D] = total 25-hydroxyvitamin D in mol/L= [DVDBP] + [DAlb] + [DFree]
[Albumin]= total serum albumin in mol/L
[Total VDBP]= total vitamin D binding protein in mol/L
[Free D] = concentration of free (unbound) 25(OH)D in mol/L
[DAlb] = concentration of albumin-bound 25(OH)D
[Bio D] = concentration of bioavailable-25(OH)D in mol/L=[Free D] + [DAlb]
[DVDBP] = concentration of VDBP-bound 25(OH)D
[UnVDBP] = free, unbound VDBP
KAlb = affinity constant between 25(OH)D and albumin = 6 x 105 M-1
KVDBP = affinity constant between 25(OH)D and VDBP = 7 x 108 M-1
EQUATIONS
Total-25(OH)D
[Total D] = concentration of total-25(OH)D in g/L ÷ 400.5 g/mol
(Molecular weight of 25(OH)D= 400.5 g/mol)
Given that [Total D] = [Free D] + [DAlb] + [DVDBP]
thus [DVDBP] = [Total D] - [DAlb] – [Free D]
Equation 1
1
Albumin
[Albumin] = serum albumin concentration in g/L ÷ 66,430 g/mol
(Molecular weight of serum albumin= 66,430 g/mol)
[Free D] + [Albumin] ↔ [DAlb]
Albumin association constant KAlb = [DAlb] ÷ ([Free D] ⋅ [Albumin])
Thus [DAlb] = KAlb ⋅ [Albumin] ⋅ [Free D]
Equation 2
(As stated in Powe et al. J Bone Min Res 2011: [Albumin] in this example denotes the
concentration of free non-vitamin bound albumin. However, given the low affinity between
albumin and vitamin D, the concentrations of total albumin and unbound albumin are effectively
equivalent ([Total Albumin] ≈ [Albumin]). As a result, [Albumin] may be estimated accurately
by measurements of total serum albumin.)
VDBP
[Total VDBP] = concentration of serum VDBP in g/L ÷ 58,000 g/mol
(Molecular weight of VDBP=58,000 g/mol)
Given that [Free D] + [UnVDBP] ↔ [DVDBP]
And VDBP association constant KVDBP = [DVDBP] ÷ ([UnVDBP] ⋅ [Free D])
Thus [Free D] = [DVDBP] ÷ KVDBP ÷ [UnVDBP]
Equation 3
Since [Total VDBP] = sum of bound and unbound VDBP = [UnVDBP] + [DVDBP]
Therefore [UnVDBP] = [Total VDBP] – [DVDBP]
Equation 4
Free-25(OH)D
From Eq. 3 and 4:
[Free D] = [DVDBP] ÷ KVDBP ÷ ([Total VDBP] – [DVDBP])
Equation 5
2
Substituting Eq. 1 into Eq. 2:
[DVDBP] = [Total D] – (KAlb ⋅ [Albumin] +1) ⋅ [Free D]
Equation 6
Substituting Eq. 6 into Eq. 5 produces the following:
[Free D] = {[Total D] – (KAlb ⋅ [Albumin] +1) ⋅ [Free D]} ÷ KVDBP ÷ ([Total VDBP] – {[Total D]
– (KAlb ⋅ [Albumin] +1) ⋅ [Free D]})
As stated in Powe et al. J Bone Min Res 2011: The equation is now limited to known constants
(KVDBP and KAlb), measured values ([Total VDBP], [Albumin], and [Total D]) and the dependent
variable for free vitamin D [Free D]. After propagating products and several rearrangements, this
can be simplified to fit the form of a second-degree polynomial:
ax2 + bx + c = 0
Where x = [Free D] = the concentration of free-25(OH)D
a = KVDBP ⋅ KAlb ⋅ [Albumin] + KVDBP
b = KVDBP ⋅ [Total VDBP] – KVDBP ⋅ [Total D] + KAlb ⋅ [Albumin] +1
c = -[Total D]
This polynomial may be solved for [Free D] using the quadratic equation:
[Free D] = -b+√𝑏 2 − 4𝑎𝑐] ÷ 2a
Bioavailable-25(OH)D
After solving for free-25(OH)D, Equation 2 can be used to calculate the concentration of
bioavailable-25(OH)D (non-VDBP bound 25(OH)D):
[Bio D] = [Free D] + [DAlb] = (KAlb ⋅ [Albumin] + 1) ⋅ [Free D]
Equation 7
3