Trypsin Assay Calculation - Please Help me!!Really in urgent (Sep/08/2008 )
Result:
1. Original Trypsin: 3 mg/ml of water
2. BAEE (1 mM) in 20 mM CaCl2, 50 mM Tris-Cl, pH 8.0
3. After purification, 1 ml Trypsin at abs 280 nm = 0.273 (0.176 at 253 nm), next fractionn at 280 nm - 0.579 (0.339 at 253 nm)
Procedures:
1. Pipette 0.98 ml BAEE solution into cuvette
2. add 0.02 ml 10 mM Tris-Cl, pH 7.0 or fraction
3. Mix Gently, start timer
4, at time = 5, read abs at 253 nm , using 10 mM Tis-Cl, pH 7.0 as blank
Dear ALL,
is there enough information for calculation on enzyme Kinetic??
THanks, PLs help!!!
1. Original Trypsin: 3 mg/ml of water
2. BAEE (1 mM) in 20 mM CaCl2, 50 mM Tris-Cl, pH 8.0
3. After purification, 1 ml Trypsin at abs 280 nm = 0.273 (0.176 at 253 nm), next fractionn at 280 nm - 0.579 (0.339 at 253 nm)
Procedures:
1. Pipette 0.98 ml BAEE solution into cuvette
2. add 0.02 ml 10 mM Tris-Cl, pH 7.0 or fraction
3. Mix Gently, start timer
4, at time = 5, read abs at 253 nm , using 10 mM Tis-Cl, pH 7.0 as blank
Dear ALL,
is there enough information for calculation on enzyme Kinetic??
THanks, PLs help!!!
why do you measure at 280 nm when the procedure tells you to measure at 253 nm?
what do you want to measure anyway? the BAEE value? apparently you have a BAEE standard already. is it not labeled?
do you measure the BAEE units of trypsin or of a trypsin inhibitor?
seems like the procedure tells you to measure only once after 5 unidentified units of time (minutes i guess) so why do you have two abs values?
and whats "next fraction"? is there a semipreparative LC involved you have not told us about?
i want to help but please work out the details
So sorry.. Maybe I need to explain more.
Actually, I need to separate three proteins from the sample (Trypsin, cytochrome c and BSA, 3 mg each/ml of water), after separation, I collect the 2 fractions (1 ml / each in 10 mM Tris-Cl Buffer, pH 7.0) wtih the highest value of absorbance at 280 nm. The absorbance values are 0.273 and 0.579. THen I conduct the trypsin assay by using BAEE (1 mM in 20 mM CaCl2, 50 mM Tris-Cl. pH 8.0)
I use 0.98 ml BAEE+ 0.02 ml Tris-Cl pH 7.0 as blank, then I add 0.02 ml of two fractions (mentioned above) in two 0.98 ml BAEE respectively.
The final results are 1. Fraction with 0.273 abs at 280 nm shows 0.176 abs at 253 nm
2. Fraction with 0.2579 abs at 280 nm shows 0.339 abs at 253 nm
I would like to know how can I conduct the conclusion on enzyme kinetic??CAn i get any information on above result??
Actually, I need to separate three proteins from the sample (Trypsin, cytochrome c and BSA, 3 mg each/ml of water), after separation, I collect the 2 fractions (1 ml / each in 10 mM Tris-Cl Buffer, pH 7.0) wtih the highest value of absorbance at 280 nm. The absorbance values are 0.273 and 0.579. THen I conduct the trypsin assay by using BAEE (1 mM in 20 mM CaCl2, 50 mM Tris-Cl. pH 8.0)
I use 0.98 ml BAEE+ 0.02 ml Tris-Cl pH 7.0 as blank, then I add 0.02 ml of two fractions (mentioned above) in two 0.98 ml BAEE respectively.
The final results are 1. Fraction with 0.273 abs at 280 nm shows 0.176 abs at 253 nm
2. Fraction with 0.2579 abs at 280 nm shows 0.339 abs at 253 nm
I would like to know how can I conduct the conclusion on enzyme kinetic??CAn i get any information on above result??
ok that makes things clearer
im sorry but i fear the answer is NO
because for the trypsin assay you dont evaluate the abs at 253 nm but the increase of abs at 253 nm over several minutes.
like you wrote in your first post:
4, at time = 5, read abs at 253 nm , using 10 mM Tis-Cl, pH 7.0 as blank
sounds like mix it, measure, wait 5 min, measure again to me but i also have heard of people measuring 253 nm every minute to reduce noise and to find out when the linear increase is over
this increase is then taken to calculate the amount of trypsin
1 BAEE unit is the amount of enzyme that causes a increase in absorbance at 253 nm of 0.003 per minute
but since you only measured 253 nm once for each fraction you can not do this.
Actually, I need to separate three proteins from the sample (Trypsin, cytochrome c and BSA, 3 mg each/ml of water), after separation, I collect the 2 fractions (1 ml / each in 10 mM Tris-Cl Buffer, pH 7.0) wtih the highest value of absorbance at 280 nm. The absorbance values are 0.273 and 0.579. THen I conduct the trypsin assay by using BAEE (1 mM in 20 mM CaCl2, 50 mM Tris-Cl. pH 8.0)
I use 0.98 ml BAEE+ 0.02 ml Tris-Cl pH 7.0 as blank, then I add 0.02 ml of two fractions (mentioned above) in two 0.98 ml BAEE respectively.
The final results are 1. Fraction with 0.273 abs at 280 nm shows 0.176 abs at 253 nm
2. Fraction with 0.2579 abs at 280 nm shows 0.339 abs at 253 nm
I would like to know how can I conduct the conclusion on enzyme kinetic??CAn i get any information on above result??
ok that makes things clearer
im sorry but i fear the answer is NO
because for the trypsin assay you dont evaluate the abs at 253 nm but the increase of abs at 253 nm over several minutes.
like you wrote in your first post:
4, at time = 5, read abs at 253 nm , using 10 mM Tis-Cl, pH 7.0 as blank
sounds like mix it, measure, wait 5 min, measure again to me but i also have heard of people measuring 253 nm every minute to reduce noise and to find out when the linear increase is over
this increase is then taken to calculate the amount of trypsin
1 BAEE unit is the amount of enzyme that causes a increase in absorbance at 253 nm of 0.003 per minute
but since you only measured 253 nm once for each fraction you can not do this.
Many Thanks!!!
I want to like how can I make the conclusion from the above result (Relative trypsin Activity)??
Also If i really want to find out the enyzme kinetic by this method, how can I make the improvements?
Thanks again!!!