Green Light Lab Report
1. Describe the mechanism by which green fluorescent protein (GFP) is able to emit fluorescent light and why ultraviolet light is required to visualise it. (5 marks)
When GFP is hit with UV light, the chromophore is hit by a photon. This changes the chromophore from ground state (A) to A*, which is a highly excitable state. Due to such a highly excitable state not being able to remain so for very long, the A* state chromophore emits a proton, lowering its state to I*, the energetic I. This I* state chromophore lowers its energy by emitting a photon of green light, lowering its energy to I. However, since I and A have such similar energies, I can convert to A and the photon can simply start the whole process again.
2. GFP is shown below …show more content…
in a cartoon ribbon visualisation form and is described as a beta barrel structure. GFP is composed of 238 amino acids and has a molecular weight of 26.9 kDa. What is the molarity of a 10 mg/mL solution of pure GFP? (3 marks)
To convert KiloDaltons of GFP into Daltons :
26.9 kDa * 1000 = 26900 Daltons (GFP)
Equation to work out molarity:
Mass / Mr = Molarity
Molarity in 1 ml = 10 / 26.900 = 3.717 * 10-4 M
.
The molarity of GFP in a 10 mg/ml solution of pure GFP would be 3.717 * 10-4.
How many molecules of GFP would there be (Avogadro’s Number = 6.023 x 1023 molecules per mole) in a 1 mL sample? (2 marks)
Avogadro’s number = molecules per mole.
Mass = Moles * Molarity
Moles = Mass / Molarity
Moles = 10 / 3.717*10-4 = 26903.4 moles.
Number of molecules = 26903.4 * 6.023*1023 = 1.6204*1028 molecules of GFP.
Methods:
3. In the laboratory session the expression of GFP was under the control of an arabinose-responsive promoter. Comment on how this control system is likely to operate. (4 marks)
In order for the arabinose-responsive promoter to express GFP, arabinose (presumably, a sugar) must be present. The arabinose-responsive promoter would regulate the expression of GFP. When there is no arabinose (presumably, like lactose a non-essential sugar for bacterial growth), there would be no GFP being produced, and with no GFP, no fluorescence.
When there is no arabinose, the arabinose-responsive promoter would be bound to PBAD. Arabinose, when present, after an hour of lag, would bind to the arabinose-responsive promoter and change its shape, promoting the binding of RNA polymerase, and as such allowing for transcription of GFP to take place. As such, GFP would be produced, until there was no more arabinose.
4. Why was the expression strain carrying an ampicillin resistance gene on the expression plasmid? (3 marks)
The expression plasmid carries an ampicillin resistance gene as a selection gene. This is because E. Coli do not normally have resistance to Ampicillin due to not possessing the resistance gene. Not all E. Coli will take up the plasmid (only around 5% will). When Ampicillin is present in media, only the E. Coli containing the plasmid with resistance will survive.
5. The culture that was provided to you in the lab session had an optical density at 600 nm of 9.8. When a 2 mL sample (prepared using a 1 mL Pasteur pipette) of this culture was centrifuged to harvest cell pellet a pellet weight of 0.0756 g was obtained using a 4-dp balance.
Would this pellet weight accurately reflect the true cell weight of the sample? Explain your answer. (3 marks)
No. As well as error associated with using small sample size and pipette, the pellet may not have just contained cells – would have contained moisture too.
Drawing on your knowledge of gravimetrical assay, comment on the accuracy of the cell weight data obtained when using the method described above. (2 marks)
It wouldn’t be accurate – moisture could be within the pellets. A way of removing the moisture, and being sure that is only the cells being measured must be done.
Suggest how the method could be developed and validated for an in-process fermentation assay (10 marks)
Fermentation is the large-scale culture of cells…
In order to measure the amount of cells that are being produced by the fermenter, the use of CO2 monitoring could be used. The amount of CO2 detected by an appropriate detector would be proportional to the amount of bacterial cells growing within the fermenter.
Could do residual testing.
Since the fermentation would be taking place on a greater scale, taking larger samples and as such, larger pieces of equipment so that there is less error taking place.
Loss on drying could work – it would remove the moisture and other impurities, leaving the cells behind. Alternatively, vaccuum filtration to draw out any moisture.
For validation, could use a non-aqueous titration such as Karl Fischer in order to determine moisture content. Also GC could be used for residual solvents.
6. Describe the methods used to lyse your cells during the lab session commenting on the function of each of the added lysis components and/or procedures (5 marks)
In order to release GFP, the bacteria (E. Coli) cell walls must be released, due to the GFP being produced in an intracellular way.
In order to lyse cells during the lab session, three methods were used:
Lysozyme – This works by using enzymes in order to assimilate the cell wall. Due to considerably weakening the cell wall, this is often paired with freezing.
Freezing in liquid nitrogen – This works by freezing bacteria. Upon the sudden dip to sub-zero temperatures, the cell wall deteriorates due to enlarged cytoplasm, causing the cell wall to burst open.
Centrifugation – This lyses bacteria cells by putting it under large centrifugal force, which separates cells from media due to differences in mass – the smaller masses being at the top and the larger masses at the bottom.
7. Describe how you would make a buffer of the following composition at a volume of 400 mL:
100 mM Tris, 2 mM EDTA, 1% v/v ,Tween20, pH 8.0
(MW Tris – 121.14 g/mol, MW EDTA = 292.24 g/mol, MW Tween20 = 1227.54 g/mol) (5 marks)
Tween20 = 4 ml
Because: 1% of 400 400 * 0.01 (if 1 = 100%, 0.01 = 1%)
Mass = Moles * Mr
Moles = molarity * volume
121.14 = Molecular weight
100 mM = miliMolar mass
Moles = 0.1 M * (400ml/1000ml) = 0.4
Mass = 0.04 moles * 121.14 = 4.8456g
Henderson-Hasselbalch Equation: pH = pKa + log[A-/HA]
Since the pKa of Tris is 8.12, if the pH was 8 then:
8 = 8.12 + log[A-/HA]
8 – 8.12 = log [A-/HA]
-0.12 = log [A-/HA]
10-0.12 = 0.759 = [A-/HA]
[A-/HA] = ([Base] – [Acid]) / [Acid] =0.759
[Acid] = [Base] / (0.759 + 1)
[Acid] = [100 mM] / (0.759 + 1) = 56.85 mM
56.85mM = 0.05685M HCl
[0.5 M HCl] / [0.05685M HCl] = 8.7950747581
400ml / 8.7950747581 = 45.48ml HCl
2 mM EDTA =
Moles = 0.002 M * (400/1000) = 0.0008
Mass = 0.0008 * 292.24 = 0.233792g
pKa (EDTA) = 2.0, 2.7, 6.2, 10.3 pI (EDTA) = ¼ (2+2.7+6.2+10.3) = 5.3 pI is the pH where there is no net charge attributed to a molecule – 50% ionised, 50% un-ionised. pI = 5.3
KI = 10-5.3 = 5.01187234E-6
8 = 5.3 + log[A-/HA]
8 – 5.3 = log [A-/HA]
2.7 = log [A-/HA]
10(2.7) = 501.2 = [A-/HA]
([Base] – [Acid]) / [Acid] = 501.2
[Acid] = [Base] / (501.2 + 1)
[2 mM] = [Base] / (501.2 + 1)
[Base] = [2 mM] * (501.2 + 1) = 1004.4 mM NaOH
1004.4 mM = 1.0044 M NaOH
[2 M] / [1.0044 M] = 1.99123855
400 ml / 1.99123855 = 200.88 ml
Buffer = 200.88 ml NaOH + 45.48ml HCl + 4ml Tween 20 = 250.36
400 – 250.36 = 149.64 left. 149.64 distilled water
8. The columns used in your lab session had a bed height of 27 mm and a diameter of 10 mm. What was the column resin volume? (5 marks)
Column resin volume = πR2h
Diameter = 10 mm Radius = 5 mm
Height = 27 mm π * 52 * 27 = 2120.6 µl (1dp) = 2.1206 ml 2.1 ml
9. If the same resin was packed to a bed height of 150 mm in a column with a 16 mm diameter, what volumetric flow rate would be required to give a 150 cm/h flow rate (5 marks).
150 = V/π x R2
150/R2 x π = V ( volumetric flow rate)
150/(82 x π) = 7.36 mL/h
10. The three elution buffers provided to you were at the following concentrations:
E1 = 0 mM Salt E2 = 100 mM Salt E3 = 500 mM Salt
a) Suggest, with reasoning a suitable salt for inclusion into this buffer (2 marks)
b) Comment, with reasoning on how you would anticipate GFP to elute off the columns using each of these buffers (4 marks)
a) (NH4)2SO4 – Mild, not to mention it’s a fairly common choice. It washes away any proteins from the column that aren’t strongly bound. As such, strongly bound (hydrophobic) proteins stay on the column.
b) E1 – Would be best at eluting GFP off of column. E2 – Would be better than E3, but worse than E1 E3 – Would be worst at eluting GFP off of column.
When eluting from the column, the lowest concentration of salt is optimal. This is because the lower salt concentration alters GFP shape due to the protein being solvated – the hydrophilic associations exposing hydrophilic parts of the protein. This, in turn makes the makes less hydrophobic interactions occur, decreasing association with the column - therefore being washed away. In contrast, a higher salt concentration would make a better binding solution.
11. Comment the value of using gradient elution profile for you samples. (3 marks)
Gradient elution would be beneficial, as everything would elute from the column. Therefore, compounds that are more highly retained would elute more quickly than in an isocratic mode.
12. The functional groups used on the HIC columns in the lab session are shown below:
Comment on the function of each of the three identified functional groups (5 marks) t-Butyl – Is hydrophobic. Will ‘attract’ hydrophobic parts of molecules.
Carboxyl – Is hydrophilic – will attract charged/strongly hydrophilic parts of molecules
Methyl – Is hydrophobic, but not as hydrophobic as t-Butyl. Will attract moderately hydrophobic molecules.
Results and Discussion:
13. Detail your observations from the GFP purification lab session from lysed cells through to GFP elution on the HIC column describing how you could incorporate in-process analytical techniques to characterise the extraction and purification process (10 marks)
During step five of the experiment, tubes 2 (pellet), 4 (whole culture 2), 5(whole culture 3) and 6(whole culture 4) all displayed fluorescence after transferring the supernatant from tube 2 to tube 3, discarding the supernatant from tubes 4, 5 and 6. This is because GFP was contained within the cells.
In step 6, tubes 4, 5 and 6 all displayed fluorescence, although to a lesser degree since the solution had been diluted. The green fluorescence was distributed throughout the tube since the pellet had been re-suspended.
With step 10, tubes 4, 5 and 6 fluoresced but once again it was a much lighter green.
In step 12, for all three columns, there was green fluorescence, however the supernatant was not green.
This appeared to stay the same during step 13, with the columns still displaying green fluorescence, whereas there was no green within the tubes.
In step 16, the columns all had a small amount of green fluorescence at the bottom, with tube 4 still not displaying any fluorescence, tube five displaying a faint amount of green fluorescence with tube six having slightly brighter fluorescence.
In process analytical techniques –
Could have used filtration for purification.
Reverse phase HPLC in order to see how well the extraction went.
Affinity chromatography
Ultrafiltration/diafiltration – for purification, however depends on how large GFP is in comparison to other proteins within cell.
14. Discuss the results obtained in the CE and HPLC analysis lab – Week 18 (5 marks). Describe some other possible downstream analysis techniques that could be used to characterise the GFP protein generated in the lab session (10 marks)
CE lab – No optimisation and validation took place within the CE lab due to speed of CE.
The above diagram is from the protein test mix. There is high signal to noise, making it seem as if there aren’t any peaks – there may well be, but any peaks that are there don’t reach above the limit of quantification. At around 2 minutes, there appears to be signals just slightly above the signal-to-noise which could be BSA.
The BSA protein peak elutes at 2.001 minutes, appearing as two peaks in one. Once again, there is signal to noise, but not as bad as the protein test mix above.
The myoglobin data shown above again shows less signal-to-noise than the previous graphs. In addition, the peak at around 1.483 is myoglobin.
This combined protein mix data below clearly shows the presence of both BSA and Myoglobin.
The graph above is from the Elutions from sample 4. High signal to noise. Peaks at 1.2, 1.5 and possible peak at around 2, which doesn’t achieve limit of quantification. Likely to be GFP, myoglobin and BSA respectively.
The graph above is from sample 5 elutions. There are peaks at 1.4, 1.7 and 2. Once again, this is likely to be GFP, myoglobin and BSA. There is however, a higher area at the significant peaks, indicating a higher concentration within the sample 5 elution, although only slightly higher. This would make sense as sample 4 was a control with no salt content; however sample 5 has a higher salt content (100mM) which would mean that more GFP would have bound to the HIC column. As such, there would be more to elute.
The graph above is from sample 6 elute. As with the other three, there are peaks at 1.1, 1.5 and 2, which correspond to GFP, myoglobin and BSA. Similarly to 5, the area of the three proteins has increased, only more significantly. Again, this makes sense since 6 had a higher salt content (500mM) meaning that more GFP would have bound to the HIC column.
Describe some other possible downstream analysis techniques that could be used to characterise the GFP protein generated in the lab session (10 marks)
First of all, downsteam analytical techniques refer to the position on DNA/RNA – upstream would be closer to 5’ end whereas downstream would be closer to the 3’ end of a coding strand.
A fluorescence assay could be used in the characterisation of GFP produced. Since signal is proportional to concentration, it could first be determined how much GFP is within the sample (assuming that it is the only fluorescent material within the sample), and then it could be used in order to calculate how much of the sample is of the wild type, since random mutation may alter the type of GFP produced.
In order to characterise the size of GFP, Gel/Capillary electrophoresis may be used. Also, so could Size Exclusion Chromatography. This could be especially useful seeing as GFP tend to form dimers (54 kDaltons).
In order to characterise charge, Capillary electrophoresis may be used.
In order to characterise the amount, qPCR and Flow cytometry may be useful
Affinity chromatography could be used.
MS, in order to characterise size, any fragmentation, detection, etc.
15. Discuss the following design of experiments-based approach to optimising a HIC-based chromatography step. In your discussion comment on
i) The use of small 300 µL columns and the more traditional 19.6 mL Tricorn columns with respect to the pros and cons of each column type (5 marks) ii) The best conditions for running this HIC process to obtain the highest yield possible (5 marks)
Figure 1: Contour plots for the recovery in percent (see labels within each contour plot) of Bovine serum albumin on (A) PreDictor plates and (B) Tricorn column.
Recovery is plotted as a function of salt concentration (y axis) and buffer ionic strength (BIS; x axis, running from 0.05 to 0.30 M) at three different pH values for the two salt types. Experimental data points are shown as black dots.
Firstly, the small 300 microliter predictor columns often give higher recoveries than the Tricorn columns, with the minimum value of 66% recovery, even giving good recoveries using smaller salt concentrations and buffer ionic strength, which Tricorn cannot, particularly struggling during pH 5.75, when almost a third of the contour plots gave recoveries 40% or below. In addition, there were more samples taken when using the Predictor columns (9 samples per contour plot), with Tricorn columns only using 5 samples per countour plot.
With both columns, as the pH increased, so did the proportion of high percentage recoveries (>80%). Similarly (in general), as the buffer strength increased, so did recoveries, as well as increased recovery rates with higher concentrations of salt. In addition, both columns found that salt 1 recoveries were higher than salt
2.
However, a very serious question of accuracy is raised over the Predictor plates, as they give up to 112% recoveries, which is not possible.
To obtain the highest yield, salt 1 should be used on predictor plates, using high BIS (M), and high salt concentrations at pH 6.75.
When GFP is hit with UV light, the chromophore is hit by a photon. This changes the chromophore from ground state (A) to A*, which is a highly excitable state. Due to such a highly excitable state not being able to remain so for very long, the A* state chromophore emits a proton, lowering its state to I*, the energetic I. This I* state chromophore lowers its energy by emitting a photon of green light, lowering its energy to I. However, since I and A have such similar energies, I can convert to A and the photon can simply start the whole process again.
2. GFP is shown below …show more content…
in a cartoon ribbon visualisation form and is described as a beta barrel structure. GFP is composed of 238 amino acids and has a molecular weight of 26.9 kDa. What is the molarity of a 10 mg/mL solution of pure GFP? (3 marks)
To convert KiloDaltons of GFP into Daltons :
26.9 kDa * 1000 = 26900 Daltons (GFP)
Equation to work out molarity:
Mass / Mr = Molarity
Molarity in 1 ml = 10 / 26.900 = 3.717 * 10-4 M
.
The molarity of GFP in a 10 mg/ml solution of pure GFP would be 3.717 * 10-4.
How many molecules of GFP would there be (Avogadro’s Number = 6.023 x 1023 molecules per mole) in a 1 mL sample? (2 marks)
Avogadro’s number = molecules per mole.
Mass = Moles * Molarity
Moles = Mass / Molarity
Moles = 10 / 3.717*10-4 = 26903.4 moles.
Number of molecules = 26903.4 * 6.023*1023 = 1.6204*1028 molecules of GFP.
Methods:
3. In the laboratory session the expression of GFP was under the control of an arabinose-responsive promoter. Comment on how this control system is likely to operate. (4 marks)
In order for the arabinose-responsive promoter to express GFP, arabinose (presumably, a sugar) must be present. The arabinose-responsive promoter would regulate the expression of GFP. When there is no arabinose (presumably, like lactose a non-essential sugar for bacterial growth), there would be no GFP being produced, and with no GFP, no fluorescence.
When there is no arabinose, the arabinose-responsive promoter would be bound to PBAD. Arabinose, when present, after an hour of lag, would bind to the arabinose-responsive promoter and change its shape, promoting the binding of RNA polymerase, and as such allowing for transcription of GFP to take place. As such, GFP would be produced, until there was no more arabinose.
4. Why was the expression strain carrying an ampicillin resistance gene on the expression plasmid? (3 marks)
The expression plasmid carries an ampicillin resistance gene as a selection gene. This is because E. Coli do not normally have resistance to Ampicillin due to not possessing the resistance gene. Not all E. Coli will take up the plasmid (only around 5% will). When Ampicillin is present in media, only the E. Coli containing the plasmid with resistance will survive.
5. The culture that was provided to you in the lab session had an optical density at 600 nm of 9.8. When a 2 mL sample (prepared using a 1 mL Pasteur pipette) of this culture was centrifuged to harvest cell pellet a pellet weight of 0.0756 g was obtained using a 4-dp balance.
Would this pellet weight accurately reflect the true cell weight of the sample? Explain your answer. (3 marks)
No. As well as error associated with using small sample size and pipette, the pellet may not have just contained cells – would have contained moisture too.
Drawing on your knowledge of gravimetrical assay, comment on the accuracy of the cell weight data obtained when using the method described above. (2 marks)
It wouldn’t be accurate – moisture could be within the pellets. A way of removing the moisture, and being sure that is only the cells being measured must be done.
Suggest how the method could be developed and validated for an in-process fermentation assay (10 marks)
Fermentation is the large-scale culture of cells…
In order to measure the amount of cells that are being produced by the fermenter, the use of CO2 monitoring could be used. The amount of CO2 detected by an appropriate detector would be proportional to the amount of bacterial cells growing within the fermenter.
Could do residual testing.
Since the fermentation would be taking place on a greater scale, taking larger samples and as such, larger pieces of equipment so that there is less error taking place.
Loss on drying could work – it would remove the moisture and other impurities, leaving the cells behind. Alternatively, vaccuum filtration to draw out any moisture.
For validation, could use a non-aqueous titration such as Karl Fischer in order to determine moisture content. Also GC could be used for residual solvents.
6. Describe the methods used to lyse your cells during the lab session commenting on the function of each of the added lysis components and/or procedures (5 marks)
In order to release GFP, the bacteria (E. Coli) cell walls must be released, due to the GFP being produced in an intracellular way.
In order to lyse cells during the lab session, three methods were used:
Lysozyme – This works by using enzymes in order to assimilate the cell wall. Due to considerably weakening the cell wall, this is often paired with freezing.
Freezing in liquid nitrogen – This works by freezing bacteria. Upon the sudden dip to sub-zero temperatures, the cell wall deteriorates due to enlarged cytoplasm, causing the cell wall to burst open.
Centrifugation – This lyses bacteria cells by putting it under large centrifugal force, which separates cells from media due to differences in mass – the smaller masses being at the top and the larger masses at the bottom.
7. Describe how you would make a buffer of the following composition at a volume of 400 mL:
100 mM Tris, 2 mM EDTA, 1% v/v ,Tween20, pH 8.0
(MW Tris – 121.14 g/mol, MW EDTA = 292.24 g/mol, MW Tween20 = 1227.54 g/mol) (5 marks)
Tween20 = 4 ml
Because: 1% of 400 400 * 0.01 (if 1 = 100%, 0.01 = 1%)
Mass = Moles * Mr
Moles = molarity * volume
121.14 = Molecular weight
100 mM = miliMolar mass
Moles = 0.1 M * (400ml/1000ml) = 0.4
Mass = 0.04 moles * 121.14 = 4.8456g
Henderson-Hasselbalch Equation: pH = pKa + log[A-/HA]
Since the pKa of Tris is 8.12, if the pH was 8 then:
8 = 8.12 + log[A-/HA]
8 – 8.12 = log [A-/HA]
-0.12 = log [A-/HA]
10-0.12 = 0.759 = [A-/HA]
[A-/HA] = ([Base] – [Acid]) / [Acid] =0.759
[Acid] = [Base] / (0.759 + 1)
[Acid] = [100 mM] / (0.759 + 1) = 56.85 mM
56.85mM = 0.05685M HCl
[0.5 M HCl] / [0.05685M HCl] = 8.7950747581
400ml / 8.7950747581 = 45.48ml HCl
2 mM EDTA =
Moles = 0.002 M * (400/1000) = 0.0008
Mass = 0.0008 * 292.24 = 0.233792g
pKa (EDTA) = 2.0, 2.7, 6.2, 10.3 pI (EDTA) = ¼ (2+2.7+6.2+10.3) = 5.3 pI is the pH where there is no net charge attributed to a molecule – 50% ionised, 50% un-ionised. pI = 5.3
KI = 10-5.3 = 5.01187234E-6
8 = 5.3 + log[A-/HA]
8 – 5.3 = log [A-/HA]
2.7 = log [A-/HA]
10(2.7) = 501.2 = [A-/HA]
([Base] – [Acid]) / [Acid] = 501.2
[Acid] = [Base] / (501.2 + 1)
[2 mM] = [Base] / (501.2 + 1)
[Base] = [2 mM] * (501.2 + 1) = 1004.4 mM NaOH
1004.4 mM = 1.0044 M NaOH
[2 M] / [1.0044 M] = 1.99123855
400 ml / 1.99123855 = 200.88 ml
Buffer = 200.88 ml NaOH + 45.48ml HCl + 4ml Tween 20 = 250.36
400 – 250.36 = 149.64 left. 149.64 distilled water
8. The columns used in your lab session had a bed height of 27 mm and a diameter of 10 mm. What was the column resin volume? (5 marks)
Column resin volume = πR2h
Diameter = 10 mm Radius = 5 mm
Height = 27 mm π * 52 * 27 = 2120.6 µl (1dp) = 2.1206 ml 2.1 ml
9. If the same resin was packed to a bed height of 150 mm in a column with a 16 mm diameter, what volumetric flow rate would be required to give a 150 cm/h flow rate (5 marks).
150 = V/π x R2
150/R2 x π = V ( volumetric flow rate)
150/(82 x π) = 7.36 mL/h
10. The three elution buffers provided to you were at the following concentrations:
E1 = 0 mM Salt E2 = 100 mM Salt E3 = 500 mM Salt
a) Suggest, with reasoning a suitable salt for inclusion into this buffer (2 marks)
b) Comment, with reasoning on how you would anticipate GFP to elute off the columns using each of these buffers (4 marks)
a) (NH4)2SO4 – Mild, not to mention it’s a fairly common choice. It washes away any proteins from the column that aren’t strongly bound. As such, strongly bound (hydrophobic) proteins stay on the column.
b) E1 – Would be best at eluting GFP off of column. E2 – Would be better than E3, but worse than E1 E3 – Would be worst at eluting GFP off of column.
When eluting from the column, the lowest concentration of salt is optimal. This is because the lower salt concentration alters GFP shape due to the protein being solvated – the hydrophilic associations exposing hydrophilic parts of the protein. This, in turn makes the makes less hydrophobic interactions occur, decreasing association with the column - therefore being washed away. In contrast, a higher salt concentration would make a better binding solution.
11. Comment the value of using gradient elution profile for you samples. (3 marks)
Gradient elution would be beneficial, as everything would elute from the column. Therefore, compounds that are more highly retained would elute more quickly than in an isocratic mode.
12. The functional groups used on the HIC columns in the lab session are shown below:
Comment on the function of each of the three identified functional groups (5 marks) t-Butyl – Is hydrophobic. Will ‘attract’ hydrophobic parts of molecules.
Carboxyl – Is hydrophilic – will attract charged/strongly hydrophilic parts of molecules
Methyl – Is hydrophobic, but not as hydrophobic as t-Butyl. Will attract moderately hydrophobic molecules.
Results and Discussion:
13. Detail your observations from the GFP purification lab session from lysed cells through to GFP elution on the HIC column describing how you could incorporate in-process analytical techniques to characterise the extraction and purification process (10 marks)
During step five of the experiment, tubes 2 (pellet), 4 (whole culture 2), 5(whole culture 3) and 6(whole culture 4) all displayed fluorescence after transferring the supernatant from tube 2 to tube 3, discarding the supernatant from tubes 4, 5 and 6. This is because GFP was contained within the cells.
In step 6, tubes 4, 5 and 6 all displayed fluorescence, although to a lesser degree since the solution had been diluted. The green fluorescence was distributed throughout the tube since the pellet had been re-suspended.
With step 10, tubes 4, 5 and 6 fluoresced but once again it was a much lighter green.
In step 12, for all three columns, there was green fluorescence, however the supernatant was not green.
This appeared to stay the same during step 13, with the columns still displaying green fluorescence, whereas there was no green within the tubes.
In step 16, the columns all had a small amount of green fluorescence at the bottom, with tube 4 still not displaying any fluorescence, tube five displaying a faint amount of green fluorescence with tube six having slightly brighter fluorescence.
In process analytical techniques –
Could have used filtration for purification.
Reverse phase HPLC in order to see how well the extraction went.
Affinity chromatography
Ultrafiltration/diafiltration – for purification, however depends on how large GFP is in comparison to other proteins within cell.
14. Discuss the results obtained in the CE and HPLC analysis lab – Week 18 (5 marks). Describe some other possible downstream analysis techniques that could be used to characterise the GFP protein generated in the lab session (10 marks)
CE lab – No optimisation and validation took place within the CE lab due to speed of CE.
The above diagram is from the protein test mix. There is high signal to noise, making it seem as if there aren’t any peaks – there may well be, but any peaks that are there don’t reach above the limit of quantification. At around 2 minutes, there appears to be signals just slightly above the signal-to-noise which could be BSA.
The BSA protein peak elutes at 2.001 minutes, appearing as two peaks in one. Once again, there is signal to noise, but not as bad as the protein test mix above.
The myoglobin data shown above again shows less signal-to-noise than the previous graphs. In addition, the peak at around 1.483 is myoglobin.
This combined protein mix data below clearly shows the presence of both BSA and Myoglobin.
The graph above is from the Elutions from sample 4. High signal to noise. Peaks at 1.2, 1.5 and possible peak at around 2, which doesn’t achieve limit of quantification. Likely to be GFP, myoglobin and BSA respectively.
The graph above is from sample 5 elutions. There are peaks at 1.4, 1.7 and 2. Once again, this is likely to be GFP, myoglobin and BSA. There is however, a higher area at the significant peaks, indicating a higher concentration within the sample 5 elution, although only slightly higher. This would make sense as sample 4 was a control with no salt content; however sample 5 has a higher salt content (100mM) which would mean that more GFP would have bound to the HIC column. As such, there would be more to elute.
The graph above is from sample 6 elute. As with the other three, there are peaks at 1.1, 1.5 and 2, which correspond to GFP, myoglobin and BSA. Similarly to 5, the area of the three proteins has increased, only more significantly. Again, this makes sense since 6 had a higher salt content (500mM) meaning that more GFP would have bound to the HIC column.
Describe some other possible downstream analysis techniques that could be used to characterise the GFP protein generated in the lab session (10 marks)
First of all, downsteam analytical techniques refer to the position on DNA/RNA – upstream would be closer to 5’ end whereas downstream would be closer to the 3’ end of a coding strand.
A fluorescence assay could be used in the characterisation of GFP produced. Since signal is proportional to concentration, it could first be determined how much GFP is within the sample (assuming that it is the only fluorescent material within the sample), and then it could be used in order to calculate how much of the sample is of the wild type, since random mutation may alter the type of GFP produced.
In order to characterise the size of GFP, Gel/Capillary electrophoresis may be used. Also, so could Size Exclusion Chromatography. This could be especially useful seeing as GFP tend to form dimers (54 kDaltons).
In order to characterise charge, Capillary electrophoresis may be used.
In order to characterise the amount, qPCR and Flow cytometry may be useful
Affinity chromatography could be used.
MS, in order to characterise size, any fragmentation, detection, etc.
15. Discuss the following design of experiments-based approach to optimising a HIC-based chromatography step. In your discussion comment on
i) The use of small 300 µL columns and the more traditional 19.6 mL Tricorn columns with respect to the pros and cons of each column type (5 marks) ii) The best conditions for running this HIC process to obtain the highest yield possible (5 marks)
Figure 1: Contour plots for the recovery in percent (see labels within each contour plot) of Bovine serum albumin on (A) PreDictor plates and (B) Tricorn column.
Recovery is plotted as a function of salt concentration (y axis) and buffer ionic strength (BIS; x axis, running from 0.05 to 0.30 M) at three different pH values for the two salt types. Experimental data points are shown as black dots.
Firstly, the small 300 microliter predictor columns often give higher recoveries than the Tricorn columns, with the minimum value of 66% recovery, even giving good recoveries using smaller salt concentrations and buffer ionic strength, which Tricorn cannot, particularly struggling during pH 5.75, when almost a third of the contour plots gave recoveries 40% or below. In addition, there were more samples taken when using the Predictor columns (9 samples per contour plot), with Tricorn columns only using 5 samples per countour plot.
With both columns, as the pH increased, so did the proportion of high percentage recoveries (>80%). Similarly (in general), as the buffer strength increased, so did recoveries, as well as increased recovery rates with higher concentrations of salt. In addition, both columns found that salt 1 recoveries were higher than salt
2.
However, a very serious question of accuracy is raised over the Predictor plates, as they give up to 112% recoveries, which is not possible.
To obtain the highest yield, salt 1 should be used on predictor plates, using high BIS (M), and high salt concentrations at pH 6.75.