Chapter 14, Problem 14.3P

Interpretation Introduction

Interpretation:

The given species has to be classified as globular or rod-like.

Concept introduction:

${\text{R}}_{\text{g}}$ Values can be calculated by using following formula,

${\text{R}}_{\text{g}}/nm=0.05690\times {\{(v_s/c{m}^3{g}^{-1})\times (M/gmo{l}^{-1})\}}^{1/3}$

Explanation of Solution

The solute particle is assumed that they are solid spheres, ${\text{R}}_{\text{g}}$ values are calculated and compared.

${\text{R}}_{\text{g}}$ Values can be calculated as follows,

${\text{R}}_{\text{g}}/nm=0.05690\times {\{(v_s/c{m}^3{g}^{-1})\times (M/gmo{l}^{-1})\}}^{1/3}$

${\text{R}}_{\text{g}}$ Values for serum albumin,

${\text{R}}_{\text{g}}/nm=0.05690\times {\{(v_s/c{m}^3{g}^{-1})\times (M/gmo{l}^{-1})\}}^{1/3}$

$\begin{array}{l}{\text{R}}_{\text{g}}/nm=0.05690\times {\{(0.752)\times (66000)\}}^{1/3}\\ {\text{R}}_{\text{g}}/nm=2.09\end{array}$

The experimental value of ${\text{R}}_{\text{g}}$ for serum albumin is 2.98, the calculated value of ${\text{R}}_{\text{g}}$ for serum albumin is 2.09.

${\text{R}}_{\text{g}}$ Values for bushy stunt virus,

${\text{R}}_{\text{g}}/nm=0.05690\times {\{(v_s/c{m}^3{g}^{-1})\times (M/gmo{l}^{-1})\}}^{1/3}$

$\begin{array}{l}{\text{R}}_{\text{g}}/nm=0.05690\times {\{(0.741)\times (10600000)\}}^{1/3}\\ {\text{R}}_{\text{g}}/nm=11.3\end{array}$

The experimental value of ${\text{R}}_{\text{g}}$ for bushy stunt virus is 12.0, the calculated value of ${\text{R}}_{\text{g}}$ for bushy stunt virus is 11.3.

${\text{R}}_{\text{g}}$ Values for DNA,

${\text{R}}_{\text{g}}/nm=0.05690\times {\{(v_s/c{m}^3{g}^{-1})\times (M/gmo{l}^{-1})\}}^{1/3}$

$\begin{array}{l}{\text{R}}_{\text{g}}/nm=0.05690\times {\{(0.556)\times (4000)\}}^{1/3}\\ {\text{R}}_{\text{g}}/nm=4.43\end{array}$

The experimental value of ${\text{R}}_{\text{g}}$ for DNA is 117.0, the calculated value of ${\text{R}}_{\text{g}}$ for DNA is 4.43.

From the values serum albumin and bushy stunt virus resembles solid sphere whereas DNA does not resembles.