1. To perform cell lysis to isolate the desired protein product (Green florescent protein)
2. To perform Gel Permeation chromatography to purify the extract

Isolation and purification of product
A. Isolation

1. 10mL of culture broth was collected in a tube.


2. The tube was centrifuged at 10,000 rmp for 5minutes.
   
3. Pellet formed at the bottom of the tube and supernatant was poured into another tube.


4. Pellet with GFP cells were confirmed by observing it under UV light.
   
   (I)Using enzyme for isolation of product
   1. Micropipette 500ul of TE buffer of pH 7.5 to resuspended the pellet.
   2. Then two drops of lysozyme was added to further resuspended the cell pellet.
   3. The enzyme was allowed to act for 15minutes.
   
   (II)Freezing and thawing
   1. The tube with pellet collected in part 1 was frozen in liquid nitrogen and then thawed with warm water.
   2. Step 1 was repeated two more times.
   
   (III)Sonication
   1. Sonication was done on ice for 4 cylce of 25 second with 10 second rest in between sonication cycle.
   2. The tube was centrifuged for 20minutes at 10,000 rpm.
   3. Pellet was collected at the bottom of the tube and was resuspended using 400uL of TE buffer.
   4. The formation of pellet with GFP cells were confirmed by observing under UV light.
   
   (IV)Purification
   1. 8 test tubes and a blank were labeled and placed in the test tube rack.
   2. The blank was filled with 2.0ml of ammonium bicarbonate (buffer). Marked the other test tubes at 2.0ml mark.
   3. The stopcock was removed and the buffer was drained from the column out into a waste beaker, till it evened out with the gel bed.
   4. Transferred extract (cell free) gently (through edges of the column) to column using a dropper.
   5. Replaced the waste beaker with the test tube labeled 1, allowed the eluant to fill the test tube to the 2.0ml mark (fraction). Continued the same process for the other 7 tubes (8 fractions altogether). The flow rate of the column was fixed at an approximate rate of 1 drop/2 seconds.
   6. Buffer was added to the column each time the buffer level was nearing the buffer matrix. The buffer matrix was not allowed to run dry at all. The buffer was added while the fractions were being collected.
   7. An additional 50ml of buffer was ran through the column after fraction 8 was collected. The column’s stopcock was screwed back.

B. Purification

1. 8 test tubes and a blank were labeled and placed in the test tube rack.
   
2. The blank was filled with 2.0ml of ammonium bicarbonate (buffer). Marked the other test tubes at 2.0ml mark.


3. The stopcock was removed and the buffer was drained from the column out into a waste beaker, till it evened out with the gel bed.


4. Transferred extract (cell free) gently (through edges of the column) to column using a dropper.


5. Replaced the waste beaker with the test tube labeled 1, allowed the eluant to fill the test tube to the 2.0ml mark (fraction). Continued the same process for the other 7 tubes (8 fractions altogether). The flow rate of the column was fixed at an approximate rate of 1 drop/2 seconds.
   
6. Buffer was added to the column each time the buffer level was nearing the buffer matrix.The buffer matrix was not allowed to run dry at all. The buffer was added while the fractions were being collected.


7. An additional 50ml of buffer was ran through the column after fraction 8 was collected.The column’s stopcock was screwed back.

Absorbance readings

Fraction	OD476
Blank	0.0
1	0.054
2	0.200
3	0.097
4	0.015
5	0.074
6	0.057
7	0.000
8	0.012

(This is how our GFP look like.nice?heez)

Further Questions:
1. Plot a graph of your A476 absorbance values (Y-axis) vs fraction number. Comment on your chromatogram.

Generally, the higher the concentration of absorbing substance, in our experiment the green fluorescent protein (GFP), the greater the amount of light that will be absorbed, that is a high absorbance reading will be reflected. From the graph above fraction 2 had the highest absorbance and this means that most of the GFP was collected in this tube. The absorbance for the first fraction is low because when the sample was first added to the column it takes time for the sample to reach the other end of the column. Therefore, the 1st tube would contain mostly ammonium bicarbonate. Starting from fraction 2, the GFP will start to elute. Therefore, the absorbance value increases to a peak and thus, this means that most of the GFP have been eluted. Finally, other fractions will have lesser GFP in it, and the absorbance decrease significantly.

2. GFP has a Mr (molecular weight) around 27,000 kD. Though we were unable to see them, the cell free extract also contained hundreds or even thousands of other proteins. Do you think a protein with a Mr of 50,000 kD would elute in a fraction before or after GFP? Why or why not?

A protein with a Mr of 50,000 kD will elute in a fraction before GFP, which has a Mr of 27,000 kD. This is because in size exclusion chromatography, as the proteins travel down the column, high molecular weight proteins will flow through faster whereas the low molecular weight proteins will spend more time interacting and diffusing into the pores of the gel resins. Hence this slows down their flow rate.

Learning points of fermentation in general

Fermentation involve in a process whereby cells convert a substrate into valuable products. There are different modes of fermentation used in industrial fermentation, for instance batch, fed-batch and continuous fermentation. Continuous fermentation is commonly used nowadays. Fermenter culture in industrial capacity often refers to highly oxygenated and aerobic growth conditions. Most industrial fermentation uses aerobic fermenters because it ensures adequate mixing and proper aeration by impeller and sparger.