Chapter 19, Problem 19.12P

(a)

Interpretation Introduction

Interpretation:

Concentration of transferrin in $\text{mg/mL}$ and iron in $\text{μg/mL}$ have to be calculated.

Concept Introduction:

Expression to compute absorbance as per Beer’s law is as follows:

  $A=\epsilon \cdot b\cdot c$

Here,

$\epsilon$ denotes molar absorptivity of species capable top absorb.

$b$ denotes path length in centimeters.

$c$ denotes concentration of species capable top absorb

$A$ denotes absorbance.

$T$ denotes transmittance.

Explanation of Solution

Expression to compute concentration as per Beer’s law is as follows:

  $c=\frac{A}{\epsilon \cdot b}$        (1)

Substitute 0.463 for $A$, ${\text{4170 M}}^{-1}{\text{cm}}^{-1}$ for $\epsilon$, and $\text{1 cm}$ for $b$ in equation (1).

  $\begin{array}{c}c=\frac{0.463}{\left({\text{4170 M}}^{-1}{\text{cm}}^{-1}\right)\left(1\text{ cm}\right)}\\ =1.11\times {10}^{-4}\text{ M}\end{array}$

Since transferrin can bind twice as much as ${\text{Fe}}^{3+}$ ion, so concentration of ${\text{Fe}}^{3+}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Fe}}^{3+}\right]=\left(2\right)\left(1.11\times {10}^{-4}\frac{\text{mol}}{\text{L}}\right)\left(\frac{55.85\text{ g}}{1\text{ mol}}\right)\left(\frac{{10}^6\text{ μg}}{1\text{ g}}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =12.39\text{mg/mL}\\ \approx 12.4\text{mg/mL}\end{array}$

Since, molar mass of transferrin is $81000\text{ g/mol}$ so concentration of transferrin is calculated as follows:

  $\begin{array}{c}\left[\text{transferrin}\right]=\left(1.11\times {10}^{-4}\frac{\text{mol}}{\text{L}}\right)\left(\frac{\text{81000 g}}{1\text{ mol}}\right)\left(\frac{{10}^3\text{ mg}}{1\text{ g}}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =8.99mg/mL\end{array}$

Thus, concentration of transferrin and iron are $8.99mg/mL$ and $12.4\text{mg/mL}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Fraction bound by transferrin has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Relation between concentration of transferrin and desferrioxamine at $470\text{ nm}$ is as follows:

  $0.424=4170c_{\text{T}}+2290c_{\text{D}}$        (2)

Relation between concentration of transferrin and desferrioxamine at $428\text{ nm}$ is as follows:

  $0.401=3540c_{\text{T}}+2730c_{\text{D}}$        (3)

Multiply equation (3) by 2290/2730.

  $0.336=2969c_{\text{T}}+2290c_{\text{D}}$        (4)

Subtract equation (4) from (2).

  $\begin{array}{c}0.424-0.336=\left(4170c_{\text{T}}+2290c_{\text{D}}\right)-\left(2969c_{\text{T}}+2290c_{\text{D}}\right)\\ 0.088=1201c_{\text{T}}\\ c_{\text{T}}=7.327\times {10}^{-5}\text{ M}\end{array}$

Substitute $7.33\times {10}^{-5}\text{ M}$ for $c_{\text{T}}$ in equation (2).

  $\begin{array}{c}0.424=4170\left(7.327\times {10}^{-5}\text{ M}\right)+2290c_{\text{D}}\\ c_{\text{D}}=5.173\times {10}^{-5}\text{ M}\end{array}$

Total concentration of ${\text{Fe}}^{3+}$ in both transferrin and desferrioxamine is calculated as follows:

  $\begin{array}{c}{\left[{\text{Fe}}^{3+}\right]}_{\text{Total}}=\left(2\right)\left(7.327\times {10}^{-5}\text{ M}\right)+5.173\times {10}^{-5}\text{ M}\\ =\text{1}\text{.98}\times {10}^{-5}\text{ M}\end{array}$

Formula to compute fraction bound by transferrin is as follows:

  ${\text{Fraction of Fe}}^{3+}\text{ in transferrin}=\frac{{\left[{\text{Fe}}^{3+}\right]}_{\text{Transferrin}}}{{\left[{\text{Fe}}^{3+}\right]}_{\text{Total}}}\left(100\text{ \%}\right)$        (5)

Substitute $\text{1}\text{.98}\times {10}^{-5}\text{ M}$ for ${\left[{\text{Fe}}^{3+}\right]}_{\text{Total}}$ and $1.46\times {10}^{-4}\text{ M}$ for ${\left[{\text{Fe}}^{3+}\right]}_{\text{Transferrin}}$ in equation (5).

  $\begin{array}{c}{\text{Fraction of Fe}}^{3+}\text{ in transferrin}=\frac{1.46\times {10}^{-4}\text{ M}}{\text{1}\text{.98}\times {10}^{-5}\text{ M}}\left(100\text{ \%}\right)\\ =73.7\text{ \%}\end{array}$

Thus, fraction bound by transferrin is $73.7\text{ \%}$.