Protein Fractionation
i) What could you conclude about the identity of your unknown protein mixture before coming
to class? Justify your conclusions.
The absorbance of each fraction was measured at 280 nm and 410 nm. At these absorbencies there
were 2 peaks, one displaying the heavier protein and the other displaying the lighter protein. Our
heavy protein has a higher absorbance at 410 nm than 280 nm, meaning that it is either
Haemoglobin or BSA. However, it was a coloured protein, which eliminates BSA. This narrows it
down to Hb which was seen to be coloured.
Our second peak is also coloured which means that it can be either Myoglobin or Cytochrome C.
No heavy protein has a pI greater than 7.5, so there is a good chance that in this sample, there is 1
heavy protein and 2 light proteins. There are 3 light proteins with a pI above 7.5, and one 1 below
7.5. From this we can conclude that we have one heavy protein and two light proteins, and each
light combination must have alpha-lactalbumin which has a pI of 4.8.
ii) How should you proceed in the second week in order to complete the purification and
identification?
The proteins are to be separated using ion exchange chromatography at pH 7.5. These proteins must
have pIs on either side of 7.5 and they are opposite charges at pH 7.5.
Zone electrophoresis is then to be used in order to separate the molecules by charge. The gel is
electrophoresed at pH 8.8, 90V for 45mins.
iii)
Protein Absorbances
2
1.8
1.6
1.4
Abs
1.2
1
Abs 280 nm
0.8
Abs 410 nm
0.6
0.4
Void
0.2
0
0
5
10
15
20
25
30
Fraction (mL)/ No. 3
Figure 1. Gel-Filtration of the protein mixture. 1.2 ml of protein mixture was loaded onto a 2.5 cm diameter
by 25cm in length Sephadex G-50 column equilibrated with buffer (50 mM Tris HCl, pH 7.5). The column was
eluted with buffer at approx. 1 ml/min, collecting 2.5 ml fractions. The absorbance of each fraction was
measured at 280 nm and 410 nm.
Cathode (-)
Group:
D1
Lane #
1
2
3
4
5
6
7
8
SDS Gel # 1
Loading order (# Protein mixture)
Protein marker
Sample 8
Sonia and Fatima
Sample 2
Rita and Stephanie
Ellen and Myat
Mik and Paul
Anna and Joseph
Anode (+)
Figure 2: SDS-PAGE, 10 μl of mixture was added to 10 μl of sample buffer containing Coomassie blue stain, was
heated for 5mins and then cooled; this was applied to a Bio-Rad precast gel. Electrophoresis was run at 200 V
for 60 min in buffer
iv) What can you conclude from these results?
Calculate the yield of any pure proteins you have isolated.
From the figures above, it can be concluded that we have 1 heavy protein and 2 light proteins. The
heavy protein was coloured and the A410 >> A280, and it moves towards the negative terminal, which
concludes that the heavy protein is Haemoglobin.
Alpha-lactalbumin protein is also present in our sample; it is the only protein with a pI lower than
7.5. The second peak was coloured, eliminating Ribonuclease, and narrowing it down to either
Cytochrome C or Myoglobin.
Yield of Hb:
E1% at 280 nm is 14.9
C=A/E1%
= 0.580/14.9
= 0.0389g/100mL
= 0.389g/L
= 0.389mg/mL x 2.5mL
= 0.9732
C=A/E1%
= 0.513/14.9
= 0.0344g/100mL
= 0.344g/L
= 0.344mg/mL x 2.5mL
= 0.8607
C=A/E1%
= 0.433/14.9
= 0.0291g/100mL
= 0.291g/L
= 0.291mg/mL x 2.5mL
= 0.7265
0.9232+0.8607+0.7265= 2.56
% yield = 2.56/4mg x 100= 64%
v) Why must you load the sample carefully in a very small volume when performing gel filtration
chromatography? Why is this not critical for the ion exchange column?
Gel-filtration chromatography separates molecules according to size and shape. The sample must be
carefully loaded to ensure that the protein mixture is properly ‘run in’ before adding running buffer
so that the sample doesn’t dilute with the buffer. It is also loaded carefully to minimise bubbling and
turbulence. Another important factor is not allowing the column to run dry. Otherwise, cracks and
fissures develop and the matrix has to be removed and repacked.
Loading the sample isn’t as critical for the ion exchange column because it separates molecules
according to charge.
vi) What will be achieved by the zone electrophoresis and the ion exchange chromatography?
Zone electrophoresis- applying an electric field across a gel at a given pH which separates proteins by
charge; proteins will migrate depending on their pI and the pH of the buffer. At the pHs on the acid
side of the pI, the protein will have a more positive charge and it will move to the cathode (-ve).The
pHs on the alkaline side of the pI will result in a protein with a negative charge, and it migrates to the
anode (+ve end).
Ion exchange chromatography- passing a mixture of proteins across a column which is covered by
positive or negative charges; some proteins will bind to the column and others will elute. This
technique also separates proteins by charge. The column is positively charged (DEAE), which means
proteins with pIs < pH of the buffer will be negatively charged and bind to the column. Opposite
charges attract. Proteins with pIs < pH of the buffer will be positively charged and will not bind to the
column but elute.
vii) What results do you predict from each procedure?
The pH of the buffer used in zone electrophoresis is 8.8. Haemoglobin has a pI of 6.9; this is lower
than the pH of the buffer, so it will be negatively charged and it will move to the anode, the positive
electrode. Alpha-lactalbumin has a pI of 4.8; this is also lower than the pH of the buffer so it will also
move towards to the anode. However, Cytochrome C has a pI of 10 and myoglobin has a pI of 7.9;
this is higher than the pH of the buffer, and will move towards the cathode; the negative electrode.
The pH of the buffer in Ion Exchange Chromatography is 7.5 and the column is positively charged
with DEAE. Alpha-lactalbumin’s pI is lower than the pH and the buffer will then be negatively
charged and it will bind to the column. Cytochrome C and Myoglobin, however, will elute from the
column because they have a pI of 10 and 7.8 respectively and the buffer will be positively charged.