
Isolation
In order to obtain Green Fluorescent Protein (GFP) which is an intracellular product, the bacteria cells need to be lysed first to release the protein.
10mL of the culture broth was collected and centrifuged at 10,000rpm for 5 minutes to obtain cells which will form a pellet at the bottom of the tube. Poured the supernatant into another tube and observed both tubes under UV light to confirm the product.
Three methods used to disrupt the cells:
Using enzymes
Freezing and thawing
SonicationStage 1
Green fluorescent protein is an intracellular product, hence the bacteria cells needed to lysed first to release the protein. Three methods of cell disruption were performed on the bacterial cells.
1. 10ml of culture broth was collected in a tube.
2. Cells were obtained by centrifuging at 10,000rpm for 5 minutes. Pellet contained the product. 3. Pellet was viewed under UV light.
Method 1 – Using Enzymes
a. Pellet was re-suspended in 500µl of TE buffer of pH 7.5 using a micropipettor.
b. 2 drops of lysozyme was added to re-suspend the pellet.
Method 2 – Freezing and Thawing
a. Tube was dipped in liquid nitrogen to freeze the suspension.
b. Cell rupturing was completed by quickly putting the tube into hot water.
c. Cell wall expands (when frozen) and contract (when thawed).
Method 3 – Sonication
a. Cell disruption is completed by the process of sonication where ultrasonic waves (of higher frequency than sound) cause the bacteria cell wall to implode under the vibration pressure.
b. Sonication on ice for 4 cycles of 25 seconds with 10 seconds of rest in between.
c. The product is in the supernatant. The cell disruption is completed by the process of sonication where ultrasonic waves cause the bacteria cell wall to implode under the vibrational pressure.
Stage 2 - Purification
Extract from stage 1 was purified by Gel Permeability Chromatography or Size Exclusion Chromatography.
This method of purification uses a column of a polymer gel resins (Sephadex G75). The resins contain very small pores in which molecules which are small enough can diffuse within.Hence, when the extract is poured into the column, the larger molecules will spend more time interacting and diffusing into the pores of the gel resins. This achieves separation of the different molecules by size.1)Label 8 test tubes and a blank
2)Filled the blank with 2mL of ammonium bicarbonate
3)Added cell free extract into the column of gel
4)Collected buffer and extract to the 2mL mark of the test tube while adding ammonium bicarbonate
5)Exposed the tubes to UV light and measured absorbance
Stage 3 – Analysis
Take absorbance readings using spectrophotometer set at 476nm.

Results for Products formed


Discussions
In this experiment, both of the graphs show us that during the 1st hr of putting in the cells to the bioreactor, the cells are at a lag phase where they are trying to adapt to the new environment which they are in. As shown in fig 1 (calculation), the OD reading of the cells are really low, which reads at 0.001.The oxygen level used was the lowest almost the whole graph seen. The pH level during the lag phrase was stable, since the cells have not started consuming any substrate and oxygen, and no waste have been produced yet.
After the 2nd hour, the cells have started to adapt to the new environment given to it. Therefore, the using of the substrate & the oxygen level start to increase which was shown in the graph Fig.2 the decreasing line of PO2 plotted by the computer showing that the amount of the oxygen level found in the bioreactor was not sufficient and air was pumped into the reactor for the production of new cells. The Abs reading was the highest. Oxygen is also one of the important factors that will affect cell growth, if the amount of oxygen needed to produce new cells is not enough, cells will not grow properly.
As cells start to multiply and use substrate, the pH inside the bioreactor will drop or increase due to the by-products of the cells and will not be at optimum for cell growth, so the reactor will automatically add in acid or base when it detects that the pH level in the reactor changes.
Foaming is also another factor with will affect the rate of the cells growth, as oxygen is sparged into the bioreactor as bubbles through the sparger & the impeller are moving at a high speed (proportional to the amount of oxygen level in the bioreactor) to provide enough oxygen to the cells, therefore foam may form, foam may cause cell to lyses when the bubbles pop. The media will be reduced by creating foam too and anti-foam is added to reduce levels of foam when the foam probe detects foam present. Most bioreactors will have baffles at the side to create a more uniform mixing by interrupting the whirlpool effect caused by stirring from the impeller.
This is called the exponential phase as it is the phase where the largest amount of oxygen level consumed by the cells for cell growth, more foaming will occur during this phase & the impeller tend to stir at a higher rate shown from the Fig 2.
During the stationary phrase after 6hr, the cell growths are already to the maximum, shown by Fig 1 (calculation) the log phase is 0 & the fig 2 of utility of the oxygen, anti-foam & the stirring speed of the impeller are more stable, no infrequently of up/down, which the cell growth are slowly going down nearly coming to the death phase.
Results for Isolation and purification of product
Absorbance reading set at 476nm.

Discussions
The graph shows the absorbance value of the GFP substance. The graph shows that fraction 3 has the highest OD reading as compared to the rest. This shows that fraction 3 has the highest concentration of GFP. When shown under UV light fraction 3 presented the highest glow intensity then fraction 4 and 6. The points after fraction 4 are off; this shows that the absorbance readings are off the decreasing scale. The points are fluctuating. Theoretically, the amount of GFP should decrease as more buffer are added through the column and eluting the GFP. But at fraction 6 in the graph the absorbance value is slightly higher then fraction 5. Most amount of GFP being eluted is in fraction 3 followed by 4 in the graph. The absorbance reading of fraction 9 in the graph is almost at 0.000; this concludes that most of the GFP has been eluted from the column.
Answers to further questions:
Sephadex G-75 contains small pore resins which do not allow large molecules to interact with. Large molecules like proteins can’t interact with Sephadex G-75, thus, a protein of Mr 50,000KD will elute first before the GFP with Mr of 27,000. Since protein are larger then GFP, it will pass through the column faster then the GFP and being eluted first. The GFP will then take its time to slowly diffuse in and interacting with the pores.