Question

This problem can be worked with Equations 19-6 on a calculator or with the spreadsheet in Figure 19-4. Transferrin is the iron-transport protein found in blood. It has a molecular mass of 81 000 and carries two${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$ions. Desferrioxamine B is a chelator used to treat patients with iron overload (see the opening of Chapter 12). It has a molecular mass of about 650 and can bind one${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$Fe31. Desferrioxamine can take iron from many sites within the body and is excreted (with its iron) through the kidneys. Molar absorptivities of these compounds (saturated with iron) at two wavelengths are given in the table. Both compounds are colorless (no visible absorption) in the absence of iron.

(a) A solution of transferrin exhibits an absorbance of 0.463 at 470 nm in a 1.000-cm cell. Calculate the concentration of transferrin in milligrams per milliliter and the concentration of bound iron in micrograms per milliliter.

(b) After adding desferrioxamine (which dilutes the sample), the absorbance at 470 nm was 0.424, and the absorbance at 428 nm was 0.401. Calculate the fraction of iron in transferrin and the fraction in desferrioxamine. Remember that transferrin binds two iron atoms and desferrioxamine binds only one.

Short answer

(a). The concentration of transferrin in milligrams 8.99mg/ml .

(b). The fraction of iron in transferrin and the fraction in desferrioxamine.26.3% .

Step-by-step solution

Calculate the concentration of transferrin:

$\mathrm{C}=\frac{\mathrm{A}}{\mathrm{eb}}$

Consider,

A as absorbance,

e as molar absorptivity,

b as pathlength.

$$\begin{gathered} \mathrm{C}=\frac{0.463}{\left[\left(4170\right)\left(1000\right)\right]} \\ \mathrm{C}=1.110\times {10}^{-4} \\ \mathrm{C}=8.99\mathrm{g}/\mathrm{mL} \end{gathered}$$

The solution for concentration of transferrin in milligrams is8.99mg/ml.

Calculate the fraction of iron in transferrin and the fraction in desferrioxamine:

$$\begin{gathered} {\mathrm{A}}_2=\sum _{}\mathrm{\epsilon cb} \\ \mathrm{At}470\mathrm{nm}:0.424=4170\left[\mathrm{T}\right]+2290\left[\mathrm{D}\right] \\ \mathrm{At}428\mathrm{nm}:0.401=3504\left[\mathrm{T}\right]+2730\left[\mathrm{D}\right] \end{gathered}$$

T is the concentration of transferrin, $\left[\mathrm{T}\right]=7.30\times {10}^{-5}\mathrm{M}$

D is the concentration of desferrioxamine,$\left[\mathrm{D}\right]=5.22\times {10}^{-5}\mathrm{M}$

The equation for fraction of Iron in transferrin is

role="math" localid="1663646842734" $\frac{2\left[\mathrm{T}\right]}{2\left[\mathrm{T}\right]+\left[\mathrm{D}\right]}=73.7\%$

To calculate the fraction of Iron in desferrioxamine

$100-73.7\%=26.3\%$