Chapter 8, Problem 10P

Summary Introduction

To evaluate: The number of copies of a 120-kd protein required in a mammalian cell and a bacterial cell in order for it to be visible as a band on an SDS gel.

Explanation of Solution

Calculation:

Given,

Concentration of protein in cells = 200mg/ mL

Mammalian cell volume = 1000 µm³

Bacterial cell volume = 1 µm³

It is assumed that 100 µg of cell extract onto a gel and 10 ng can be detected in a single band by silver staining the gel.  Thus, the calculation would be in two parts- first calculate the number of cell-equivalents that can be loaded onto gel, and then the copies of a protein that can be detected in the band.

Mammalian cell equivalents

$\begin{array}{l}=\frac{\text{100µg}}{\text{gel}}\text{×}\frac{\text{mL}}{\text{200}\text{mg}}\text{×}\frac{\text{cell}}{\text{1000}{\text{µm}}^{\text{3}}}\text{×}\frac{\text{mg}}{\text{1000}\text{µg}}\text{×}\frac{{\left({\text{10}}^{\text{4}}\text{µm}\right)}^{\text{3}}}{{\left(\text{cm}\right)}^{\text{3}}}\text{×}\frac{{\text{cm}}^{\text{3}}}{\text{mL}}\\ =\frac{\text{100µg}}{\text{gel}}\text{×}\frac{\text{mL}}{\text{200}\text{mg}}\text{×}\frac{\text{cell}}{\text{1000}{\text{µm}}^{\text{3}}}\text{×}\frac{\text{mg}}{\text{1000}\text{µg}}\text{×}\frac{{\text{10}}^7{\text{µm}}^{\text{3}}}{{\text{cm}}^{\text{3}}}\text{×}\frac{{\text{cm}}^{\text{3}}}{\text{mL}}\\ (Commonunitsinthenumeratoranddenominatorwillgetcancelled)\\ \text{= 5}\text{×}{\text{10}}^{\text{5}}\text{cells/ gel}\end{array}$

Bacterial cell equivalents

$\begin{array}{l}=\frac{\text{100µg}}{\text{gel}}\text{×}\frac{\text{mL}}{\text{200}\text{mg}}\text{×}\frac{\text{cell}}{\text{1000}{\text{µm}}^{\text{3}}}\text{×}\frac{\text{mg}}{\text{1000}\text{µg}}\text{×}\frac{{\left({\text{10}}^{\text{4}}\text{µm}\right)}^{\text{3}}}{{\left(\text{cm}\right)}^{\text{3}}}\text{×}\frac{{\text{cm}}^{\text{3}}}{\text{mL}}\\ =\frac{\text{100µg}}{\text{gel}}\text{×}\frac{\text{mL}}{\text{200}\text{mg}}\text{×}\frac{\text{cell}}{{\text{1µm}}^{\text{3}}}\text{×}\frac{\text{mg}}{\text{1000}\text{µg}}\text{×}\frac{{\text{10}}^7{\text{µm}}^{\text{3}}}{{\text{cm}}^{\text{3}}}\text{×}\frac{{\text{cm}}^{\text{3}}}{\text{mL}}\\ (Commonunitsinthenumeratoranddenominatorwillgetcancelled)\\ \text{= 5}\text{×}{\text{10}}^8\text{cells/ gel}\end{array}$

Protein (120kD) copies in a 10 ng band

$\begin{array}{l}=\frac{\text{10}\text{ng}}{\text{band}}\text{×}\frac{\text{nmol}}{\text{120,000}\text{ng}}\text{×}\frac{{\text{6×10}}^{\text{14}}\text{molecules}}{\text{nmol}}\\ (Commonunitsinthenumeratoranddenominatorwillgetcancelled)\\ \text{= 5}\text{×}{\text{10}}^{10}\text{molecules/ band}\end{array}$

Thus to load 5×10⁵ mammalian cells per gel and to detect 5×10¹⁰ protein molecules in a band, there must be 10⁵ copies of protein per cell $\left(5\times {10}^{10}/5\times {10}^5\right)$. Similarly for a bacterial cell, there must be 100 copies of the protein per cell $\left(5\times {10}^{10}/5\times {10}^8\right)$ in order to be detectable as a silver-stained band on a gel.