Question

Ferritin (molecular mass 450 000), transferrin (molecular mass 80 000), and ferric citrate were separated by molecular exclusion chromatography on Bio-Gel P-300. The column had a length of 37 cm and a 1.5-cm diameter. Eluate fractions of 0.65 mL were collected. The maximum of each peak came at the following fractions: ferritin, 22; transferrin, 32; and ferric citrate, 84. (That is, the ferritin peak came at an elution volume of 22 3 0.65 5 14.3 mL)Assuming that ferritin is eluted at the interstitial volume and that ferric citrate is eluted at ${\mathbf{V}}_{\mathbf{m}}$, find${\mathbf{K}}_{\mathbf{av}}$for transferrin.

Short answer

The${\mathrm{K}}_{\mathrm{av}}$ for transferrin is${\mathrm{K}}_{\mathrm{av}}=0.16$

Step-by-step solution

Define ferric citrate:

Ferric citrate or iron (III) citrate describes any complex compounds formed by binding to any few conjugate bases derived from citric acid and ferric ions.

Calculate the maximum values for ferritin, ferric citrate, and transferrin:

Consider the following:

Ferritin maximum:

${\mathrm{V}}_{\mathrm{o}}=22\times 0.65\mathrm{mL}=14.3\mathrm{mL}$

Ferric citrate maximum:

${\mathrm{V}}_{\mathrm{m}}=84\times 0.65\mathrm{mL}=54.6\mathrm{mL}$

Transferrin maximum:

${\mathrm{V}}_{\mathrm{t}}=32\times 0.65\mathrm{mL}=20.8\mathrm{mL}$

Calculate Kv for transferrin:

Next calculate the ${\mathrm{K}}_{\mathrm{av}}$for transferrin by the following equation

$$\begin{gathered} {\mathrm{K}}_{\mathrm{av}}=\frac{{\mathrm{V}}_{\mathrm{t}}-{\mathrm{V}}_{\mathrm{o}}}{{\mathrm{V}}_{\mathrm{m}}-{\mathrm{V}}_{\mathrm{o}}} \\ {\mathrm{K}}_{\mathrm{av}}=\frac{20.8-14.3}{54.6-14.3} \\ {\mathrm{K}}_{\mathrm{av}}=0.16 \end{gathered}$$

The solution to${\mathrm{K}}_{\mathrm{av}}$ for transferrin is${\mathrm{K}}_{\mathrm{av}}=0.16$