Question

Transferrin is the iron-transport protein found in blood. It has a molecular mass of 81000 and carries two Fe³⁺ ions. Desferrioxamine B is potent iron chelator used to treat patient with iron overload. It has a molecular mass of 650 and can bind one Fe³⁺. Desferrioxamine can take iron from many sites within the body and is excreted (with its iron) through the kidneys. The molar absorptivities of these compounds are given in the following table. Both compounds are colorless (no visible absorption) in the absence of iron.

                                                               e (M^-1 cm^-1)                    

l (nm)                           Transferrin            Desferrioxamine

428                          3540                   2730

470                          4170                   2290       

  (a) A solution of transferrin has an absorbance of 0.463 at 470 nm in a 1.000 cm cell. Find the concentration of transferrin in mg/mL and the concentration of iron in μg/mL.

(b) After addition of 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. Remember that transferrin binds two Fe³⁺ ions and desferrioxamine binds only one.

Answer 1

From Beer's law we have

A=ebc

Where A is absorbance
e is the molar absorbtivity with units of L mol^-1 cm^{-1
b is the path length of the sample - that is, the path length of the cuvette in which the sample is contained. We will express this measurement in centimetres.
c is the concentration of the compound in solution, expressed in mol L-1}

0.463 = 4170 M-1 cm-1 x 1 cm x C

C=0.463/4170

C = 1.11 x 10^-4 M is the concentration of transferrin in the sample

M is in moles/litre we need mg/mL

molecular mass of transferring is 81000 g/mol

so 1.11 x 10^-4 M is 8.991 g/L or 8.991 mg/mL

concentration of transferrin in the sample is  8.991 mg/mL

Trannsferring has molecular mass of 81000 and carries two Fe³⁺ ions MW of iron is 55.85

so 81000 g/mol of Transferring has 111.7 g of iron

so 8.991 mg of transferring has 0.01239 mg of iron or 12.30 g/mL of iron

A470 = 0.424 = 4170 xC_T + 2290 x C_D

A428 = 0.401 = 3540 xC_T + 2730 x C_D

where C_T is the concentration of transferrin and C_D is the concentration of Desferrioxamine

Solving as a set of simultaneous equations

0.424 = 4170C_T + 2290C_D

0.401 = 3540C_T + 2730C_D multiply by 1.177

0.472 = 4170C_T + 3213C_D

-0.424 = -4170C_T - 2290C_D   

_{------------------------------------------}

0.048 = 923C_D

C_D = 0.048/923 = 5.20 x 10^-5M concentration of Desferrioxamine in the sample

substituting for C_D in one of the equation we can calculate C_T

0.424 = 4170C_T + 2290 x 5.20 x 10^-5

0.3049 = 4170C_T

C_T = 0.3049/4170

C_T = 7.31 x 10^-5M concentration of Transferrin in the sample

Desferrioxamine B is potent iron chelator and can take iron from Transferrin

We will first calculate the concentration of iron in Transferrin it is 7.31 x 10^-5M x 2 as one Transferring has 2 iron molecules

that is 1.46 x 10^-4M since Desferrioxamine can take up One iron per molecule of Desferrioxamine

The amount of iron taken up by Desferrioxamine is 5.20 x 10^-5M so the amount of iron left in Transferrin is 9.4 x 10^-5M

fraction of iron in transferrin is 9.4 x 10^-5M/1.46 x 10^-4M = 0.645