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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+√ð 2 − 4ðð] ÷ 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