Isolation and Purification of Acid Phosphatase<\/p>\n
Name of Student<\/p>\n
Institution Affiliation<\/p>\n
Isolation and Purification of Acid Phosphatase<\/p>\n
Results<\/p>\n
Table 1. Piece Protein Assay<\/p>\n
Fraction Number Total Volume (mL) Dilution Used Average A595 Protein Conc. Dilute Fraction (mg\/mL) *10^-3 Protein conc. Fraction (mg\/mL)<\/p>\n
*10^-3 Total Protein (mg)<\/p>\n
*10^-5<\/p>\n
1 0.05 1 Data error 1.2850 1.2850 6.425<\/p>\n
2 0.05 1 Data error 1.3037 1.3037 6.5185<\/p>\n
3 0.05 1 1153.0173 0.9451 0.9451 4.7255<\/p>\n
4 0.05 1 294.7100 0.3567 0.3567 1.7835<\/p>\n
5 0.05 1 316.9231 0.3804 0.3804 1.902<\/p>\n
3 0.05 1 1102.5018 0.9290 0.9290 4.645<\/p>\n
3 0.05 1 1136.7230 0.9401 0.9401 4.7005<\/p>\n
4 0.05 1 318.4715 0.3820 0.3820 1.91<\/p>\n
4 0.05 1 308.3141 0.3713 0.3713 1.8565<\/p>\n
5 0.05 1 225.1985 0.2797 0.2797 1.3985<\/p>\n
5 0.05 1 268.9936 0.3288 0.3288 1.644<\/p>\n
1 0.05 2 Data error 1.08911 2.17822 10.8911<\/p>\n
1 0.05 5 847.9720 0.8128 4.064 20.32<\/p>\n
1 0.05 5 875.2789 0.8281 4.1405 20.7025<\/p>\n
1 0.05 5 901.4716 0.8421 4.2105 21.0525<\/p>\n
2 0.05 5 663.0801 0.6914 3.457 17.285<\/p>\n
2 0.05 5 710.5997 0.7256 3.628 18.14<\/p>\n
2 0.05<\/p>\n
5 745.9015<\/p>\n
0.7496 3.748 18.74<\/p>\n
Table 2. Acid Phosphatase Activity<\/p>\n
Fract<\/p>\n
# Total volume<\/p>\n
ml Dilution<\/p>\n
used Volume<\/p>\n
Used in assay(mL) Avg. A405 [PNP]<\/p>\n
in<\/p>\n
Assay<\/p>\n
(M)*10^-2 Total<\/p>\n
PNP<\/p>\n
(mol)<\/p>\n
*10^-2 [PNP]<\/p>\n
Dilute Fraction<\/p>\n
(mole\/mL)<\/p>\n
*10^-2 [PNP]<\/p>\n
in<\/p>\n
Fraction<\/p>\n
(mol\/mL) Units<\/p>\n
mL Total<\/p>\n
Units<\/p>\n
1-1 0.05 40 0.1 0.3610 1.9202 7.68 1.896 0.7584 0.7584 3.07152<\/p>\n
1-2<\/p>\n
0.05<\/p>\n
40<\/p>\n
0.1<\/p>\n
0.3505<\/p>\n
1.8644<\/p>\n
7.4576<\/p>\n
1.8414<\/p>\n
0.737<\/p>\n
0.737<\/p>\n
2.98485<\/p>\n
2-1 0.05 40 0.1 0.3504 1.864 7.456 1.841 0.7364 0.7364 2.98242<\/p>\n
2-2 0.05 40 0.1 0.3488 1.8553 7.4212 1.832 0.7328 0.7328 2.96784<\/p>\n
3-1 0.05 40 0.1 0.3663 1.9484 7.7936 1.924 0.7696 0.7696 3.117<\/p>\n
3-2 0.05 40 0.1 0.3669 1.9516 7.8064 1.9275 0.771 0.771 3.12255<\/p>\n
4-1 0.05 40 0.1 0.3596 1.9128 7.6512 1.889 0.7556 0.7556 3.06018<\/p>\n
4-2 0.05 40 0.1 0.3976 2.115 8.46 2.09 0.836 0.836 3.3858<\/p>\n
5-1 0.05 40 0.1 0.3102 1.65 6.6 1.63 0.652 0.652 2.6406<\/p>\n
5-2 0.05 40 0.1 0.2970 1.58 6.32 1.56 0.624 0.624 2.5272<\/p>\n
9. Calculation of the concentration of p-nitrophenol(PNP) from the corrected average absorbance reading. <\/p>\n
A = Elc where A- Absorbance<\/p>\n
l- Path length<\/p>\n
c- Concentration<\/p>\n
E \u2013 Molar absorptivity<\/p>\n
Therefore concentration, C = A\/El<\/p>\n
= 0.3610\/(1.88 * 10^4) * 1<\/p>\n
=1.9202 * 10^-5<\/p>\n
C = A\/El<\/p>\n
= 0.3505\/(1.88 * 10^4) * 1<\/p>\n
= 1.8644 * 10^-5<\/p>\n
Repeating this for all the values yields the concentration of all the other fractions. <\/p>\n
10. The total number of moles of the PNP is a product of its concentration and the total volume of the assay mixture. The volume in this case includes the volume of potassium hydroxide which is 4.0 ml<\/p>\n
Therefore, Total moles = Concentration of PNP * Total Volume of assay<\/p>\n
For Fraction 1, Total moles = (1.9202 * 10^-5 ) * (4.0 * 1000)<\/p>\n
= 7.68 * 10^-2 moles<\/p>\n
For fraction 1-2, Total moles = (1.8644 * 10^-5) * (4.0 *1000)<\/p>\n
= 7.4576 * 10^-2 moles<\/p>\n
For fraction 2-1, Total moles = (1.864 * 10^-5) * (4.0 * 1000)<\/p>\n
= 7.456 * 10^-2 moles<\/p>\n
For fraction, 2-2, Total moles = (1.8553 * 10^-5) * 4000<\/p>\n
= 7.4212 * 106-2moles<\/p>\n
For Fraction, 3-1, Total moles = (1.9484 * 10^-5) * 4000<\/p>\n
= 7.7936 * 10^-2 moles<\/p>\n
For Fraction, 3-2, Total moles = (1.9516 * 10^-5) * 4000<\/p>\n
7.8064 * 10^-2moles<\/p>\n
For Fraction, 4-1, Total moles = (1.9128 * 10^-5) * 4000<\/p>\n
= 7.6512 * 10^-2moles<\/p>\n
For, 4-2, Total moles = (2.115 * 10^-5) * 4000<\/p>\n
= 8.46 * 10^-2 moles<\/p>\n
For, 5-1, Total moles = (1.65 * 10^-5) * 4000<\/p>\n
=6.6 * 10^-2 moles<\/p>\n
For 5-2, Total moles = (1.58 * 10^-5) * 4000<\/p>\n
= 6.32 * 106-2 moles<\/p>\n
11. moles PNP\/ml diluted fraction = Total moles \/ Amount of diluted sample<\/p>\n
= (7.68 * 10^-2)\/4.05<\/p>\n
= 1.896 * 10^-2 <\/p>\n
To get the volume of the undiluted solution, the moles PNP\/mL of diluted fraction is multiplied by the dilution factor of 40<\/p>\n
Therefore, moles of undiluted fraction = moles of diluted fraction * 40<\/p>\n
= (1.896 * 10^-2) * 40 <\/p>\n
= 0.7584<\/p>\n
12. One unit of enzyme activity produces 1mole of PNP in 15 minutes at 300c. <\/p>\n
0.7584 <\/p>\n
1 unit of enzyme activity is equivalent to 1mole of PNP<\/p>\n
Therefore 0.7584 mole is equivalent to 0.7584 units<\/p>\n
This is repeated through all the twelve samples<\/p>\n
The total number of units is the product of the units\/mole multiplied by the total volume<\/p>\n
0.7584 * 4.05 = 3.07152<\/p>\n
Table 3. Purification of Acid Phosphatase<\/p>\n
Fraction Total Volume (mL) Protein conc. (mg\/mL) Total protein (mg) Total Activity (Units) Specific Activity (units\/mg) % Recovery Purification (fold)<\/p>\n
1 4.05 1.2850 6.425 3.07152 0.4781 47.81 1.4781<\/p>\n
2 4.05 1.3037 6.5185 2.98242 0.458 45.8 0.458<\/p>\n
3 4.05 0.9451 4.7255 3.117 0.66 66 0.66<\/p>\n
4 4.05 0.3567 1.7835 3.06018 1.716 100 1<\/p>\n
5 4.05 0.3804 1.902 2.6406 1.388 100 1<\/p>\n
Discussion<\/p>\n
The acid phosphatase has been successfully purified and isolated. The specific activity obtained ha at first shown significant change in the trend and the recovery has nearly attained a specific activity of nearly 47.81%. In table 3, it is clear that with each fraction addition, there was a significant decrease in the total protein that was to be produced. This significant decrease is due to increase in the amount of unwanted proteins in each step that follows. The use of p-nitrophenol was used to test for the enzyme in order to cleave the phosphate group. In addition to that, the use of Potassium hydroxide ensured the reaction turned yellow in the long run for the observable change to be clearly visible. With each subsequent procedure, the specific activity tended to increase. The increase indicates that the process was successful and it is similar to the expected trends (Vallee). <\/p>\n
Dilution analysis was put into consideration at each step of the overall process. It was necessary to report the overall activity in the entire fraction. However, each step was not effective. It may be due to there being unintentional and excess dilution. Excess dilution would significantly lead to errors in enzyme activity. Enzyme purification is a solvent-sensitive process thus explaining the sources of the discrepancies. The results therefore indicate a successful experiment due to increase in enzyme activity from one step to the next. The 100% is not the acid phosphatase as it is expected. This may be due to the presence of solid pellets at the bottom of the equipment. <\/p>\n
Reference<\/p>\n
Vallee, B. L. (2012, December). Zinc biochemistry: a perspective. In\u00a0Biotechnological Applications of Proteins and Enzymes: Papers Presented at a Conference Honoring the Sixtieth Birthday of Professor Ephraim Katchalski-Katzir, Held at Kiryat Anavim, Israel, May 23-27, 1976\u00a0(p. 223). Elsevier.<\/p>\n","protected":false},"excerpt":{"rendered":"
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