Chapter 19, Problem 19.AE

a)

Interpretation Introduction

Interpretation:

The concentration of transferrin (in milligrams) and the concentration of bound iron in micrograms per millilitre has to be calculated.

Concept introduction:

Beer-lambert law:

$\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}$

$\begin{array}{l}\text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

To calculate concentration of transferrin (in milligrams) and the concentration of bound iron in micrograms per millilitre

Answer to Problem 19.AE

The concentration of transferrin in milligrams is $\text{8}\text{.99}\text{mg/ml}$.

The fraction of Iron is $\text{12}\text{.4}\text{μg/L}$ .

Explanation of Solution

From absorbance, the concentration can be calculated as,

$\begin{array}{l}\text{c=}\frac{\text{A}}{\text{εb}}\\ \\ \text{c=}\frac{\text{0}\text{.463}}{\left[\left(\text{4170}\right)\left(\text{1000}\right)\right]}\\ \\ \text{c=1}{\text{.110×10}}^{\text{-4}}\\ \\ \text{c=8}\text{.99}\text{g/mL}\\ \\ \text{c=8}\text{.99}\text{mg/ml}\end{array}$

The concentration of transferrin in milligrams = $\text{8}\text{.99}\text{mg/ml}$

The fraction of Iron is $\text{2}{\text{.220×10}}^{\text{-4}}\text{M=0}\text{.0124}\text{g/L=12}\text{.4}\text{μg/L}$

b)

Interpretation Introduction

Interpretation:

The fraction of Iron in transferrin and fraction of deferrioxamine has to be calculated.

Concept introduction:

Beer-lambert law:

$\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}$

$\begin{array}{l}\text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

To calculate fraction of Iron in transferrin and fraction of deferrioxamine

Answer to Problem 19.AE

The fraction of Iron in transferrin $\text{73}\text{.7\%}$ and desferrioxamine is $\text{26}\text{.3\%}$ .

Explanation of Solution

${\text{A}}_{\text{λ}}\text{=}{\displaystyle \sum \text{εcb}}$

At $\text{470}\text{nm:0}\text{.424=4170}\left[\text{T}\right]\text{+2290}\left[\text{D}\right]$

At $\text{428}\text{nm:0}\text{.401=3504}\left[\text{T}\right]\text{+2730}\left[\text{D}\right]$

Here, $\left[\text{T}\right]\text{and}\left[\text{D}\right]$ are the concentrations of transferrin and desferrioxamine.

Solve using simultaneous equations,

$\begin{array}{l}\left[\text{T}\right]\text{=7}{\text{.30×10}}^{\text{-5}}\text{M}\\ \\ \left[\text{D}\right]\text{=5}{\text{.22×10}}^{\text{-5}}\text{M}\end{array}$

The fraction of Iron in transferrin is calculated as

Fraction of Iron in transferrin $\frac{\text{2}\left[\text{T}\right]}{\text{2}\left[\text{T}\right]\text{+D}}$= $\text{73}\text{.7\%}$

The fraction of Iron in desferrioxamine is calculated as,

Fraction of Iron in desferrioxamine = $\text{100-73}\text{.7\%=26}\text{.3\%}$