A report on an experiment I did of doing electrophoresis with proteins

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• The software “ImageJ” is used to ”invert” the contrast of the image and convert this inverted picture into a
matrix of numbers which gets saves as a text file.
FIG. 13: Inverted image of the strip of the scanned image of the gel along the dilution series
• Inversion is done to make the numerical analysis more intuitive so that places in the image which are brighter
because of the dye attaching to the protein have higher numerical values.
• By looking at this picture the rectangular regions are selected in the strip which correspond to the protein spots
and a C++ program was written which takes this text file as input, stores it as a matrix and sums the numbers
in every rectangular region corresponding to each protein spot.
I. The data analysis
• So at the end of the data extraction one has a set of pairs of numbers {m i , I i } where I i is the intensity of of the
spot corresponding to m i mass of the protein and m i+1 = 1 2 m i. Using Mathematica7 the following two graphs
are plotted,
– I i vs m i
– I i vs i
• The graph I i vs m i shows somewhat of a linear behaviour if the extreme values are dropped. At the higher
mass end (i.e > 6µg) the intesity seems to stall and at the very low mass (i.e < 0.75µg) the intensity seems to
drop sharply most likely because of being very close to the lowest detection limits of this staining procedure.
At a protein mass of about 0.375µg the stain is just barely visible.
• A typical running of a dilution gel has 6 dilutions as shown in the photographs of the strips and
the intensities as defined earlier as the sum of the matrix entries in the sub-rectangles look like ,
{4.68083 × 10 6 , 3.7271 × 10 6 , 3.01742 × 10 6 , 2.69898 × 10 6 , 2.52403 × 10 6 , 2.40741 × 10 6 } corresponding to
{12µg, 6µg, 3µg, 1.5µg, 0.75µg, 0.375µg}.
This data can be analyzed in the following ways,
• One can plot the points {(1, 4.68083 × 10 6 ), (2, 3.7271 × 10 6 ), (3, 3.01742 × 10 6 ), (4, 2.69898 × 10 6 ), (5, 2.52403 ×
10 6 ), (6, 2.40741 × 10 6 )} and fit the function I = A x n to it and find the optimum value of A and n for that. The
way to think of this graph is that given the set of points {(m i , I i )|i ∈ {1, 2, ..., 6}} one is constructing from it a
set {(i, I i )|i ∈ {1, 2, ..., 6}} such that i = 1 + log 2
(
m 1
m i
). Here which is m 1 and which is m 2 etc is unambiguous
since we know precisely in the making of the dilution series that m i+1 = 1 2 m i. So m 1 is always the highest mass
sent through the gel.