Using 19F-NMR to Study the Enzyme Kinetics of Porcine Kidney Acylase I: Lab Report<\/p>\n
Name of Student<\/p>\n
Institution Affiliation<\/p>\n
USING 19F-NMR TO STUDY THE ENZYME KINETICS OF PORCINE KIDNEY ACYLASE I <\/p>\n
Results<\/p>\n
Experiment 5 Week 2.<\/p>\n
1.Reagents used were 0.0447g tfa-Gly-OH<\/p>\n
3ml 0.1M Phosphate buffer 20% <\/p>\n
pH 7.52<\/p>\n
Need 0.0078 porcine kidney acylace I<\/p>\n
Two peaks were observed, a small peak to the left (<10%) that represents any auto hydrolyzed substrate (trifluoroacetic acid) and a large peak on the right representing the intact substrate (N-tfa-Gly-OH). The F-NMR spectrum of the sample is as shown below <\/p>\n
Enzyme Peak 1 (%hydrolase) Peak 2<\/p>\n
Control 1 4.38 24.01 75.99<\/p>\n
2 6.38 35.27 64.73<\/p>\n
3 8.38 46.97 53.03<\/p>\n
4 10.38 58.53 41.47<\/p>\n
5 12.38 70.15 29.85<\/p>\n
6 14.38 80.45 19.55<\/p>\n
7 16.38 89.76 10.24<\/p>\n
8 18.38 96.98 3.02<\/p>\n
9 20.38 100 <\/p>\n
2. <\/p>\n
Enzyme Peak 1 (%hydrolase) Peak 2<\/p>\n
Control 1 4.16 15.66 84.34<\/p>\n
2 6.16 25.58 74.42<\/p>\n
3 8.16 35,86 64.14<\/p>\n
4 10.16 46.63 53.37<\/p>\n
5 12.16 57.78 42.22<\/p>\n
6 14.16 67.98 32.02<\/p>\n
7 16.16 77.24 22.74<\/p>\n
8 18.16 86.77 13.23<\/p>\n
9 20.16 94.94 5.04<\/p>\n
10 22.16 100 <\/p>\n
The table below shows the % hydrolysis versus time for each of the kinetic trials. The time at which the first spectrum of the acquisition began was at 3 minutes and 11 seconds. Afterwards, each subsequent spectrum in the array began and is recorded at two minutes interval. The times in the table below are expressed in minutes, with the seconds expressed as decimal values. <\/p>\n
Time Peak 1 Peak 2<\/p>\n
3.11 12.52 87.48<\/p>\n
5.11 21.44 78.56<\/p>\n
7.11 32.22 67.78<\/p>\n
9.11 43.16 56.84<\/p>\n
11.11 54.20 45.80<\/p>\n
13.11 64.47 35.33<\/p>\n
15.11 74.90 25.10<\/p>\n
17.11 84.10 15.90<\/p>\n
4. <\/p>\n
The slope of the graph is the rate of the reaction. <\/p>\n
Slope of peak line 1 = Change in Y\/Change in X<\/p>\n
The two points for peak line 1 are [10.4, 50], [6, 26]<\/p>\n
(Y2-Y1)\/(X2-X1)<\/p>\n
= (50 \u2013 26) \/ (10.4 \u2013 6)<\/p>\n
= 24\/4.4<\/p>\n
= 5.454545<\/p>\n
The rate of the reaction is therefore 5.4545 %hydrolysis\/min<\/p>\n
Slope for peak line 2 = Change in Y \/ Change in X<\/p>\n
= (Y2-Y1)\/(X2-X1)<\/p>\n
The two points for peak line 2 are [10.4, 50], [14, 30]<\/p>\n
= (50 \u2013 30) \/ (10.4 \u2013 14)<\/p>\n
= -5.6<\/p>\n
The rate of the reaction is therefore -5.6 %hydrolysis\/min5.<\/p>\n
Molarity = No. of Moles of Solute \/ Volume of the solution <\/p>\n
Number of moles = Mass of the solute \/ Molar mass<\/p>\n
=0.0447\/ 171.07<\/p>\n
=2.613 * 10^-4 moles<\/p>\n
To convert to micromoles, multiply by 1000000<\/p>\n
=2.613 x 10^-4 x 1000000<\/p>\n
261.3umoles<\/p>\n
Molarity = 261.3 \/ 0.002<\/p>\n
=130.65<\/p>\n
For the second sample, Number of moles = 0.0078\/171.07<\/p>\n
4.56 x 10^-5 x1000000<\/p>\n
= 45.596 umolesMolarity = 45.596 \/ 0.002<\/p>\n
= 22.798<\/p>\n
6. Concentration of the enzyme solution = Mass of enzyme\/ Volume of the enzyme<\/p>\n
= 0.0000075\/0.002<\/p>\n
=0.00375mg\/mL<\/p>\n
For week 2 Concentration = Mass of enzyme \/ Volume of the enzyme <\/p>\n
=0.0000447\/0.0025<\/p>\n
=0.01788mg\/mL<\/p>\n
For the third enzyme = 0.0000078\/0.001<\/p>\n
= 0.0078mg\/mL<\/p>\n
7. Unit activity = %hydrolysis\/min x Xumol substrate\/100%hydrolysis x 1 \/ Y mg enzyme x 60min\/1hr<\/p>\n
= 5.4545 x 261.3 x 1\/0.00078 x 60 <\/p>\n
=1.096 x 10^8<\/p>\n
For the second sample<\/p>\n
Unite activity = -5.6 x261.3 x1\/0.00078 x 60<\/p>\n
-112560000<\/p>\n
Discussion<\/p>\n
Enzyme catalysis by the enzyme acylase I of the hydrolysis of the substrate N-trifluoroacetyglycine is significantly different from that in 995D20 with that determined in 20%D2o. There are many causes of the differences due to isotope differences. There are different acylases in different tissues. There is also the isotopic acylase which hydrolyzes enzymes. The difference like the reactions is due to the nature of their structures. For instance, Acylase I is purified through homogenization from the liver in addition to the kidney and the intestines. They are found to contain similar isotopes with the same molecular mass of 45. As long as the porcine acylase is substrate specific, the relative extent of hydrolysis is different. The porcine kidney acylase is a somewhat stable enzyme at pH 7.3. The difference in the molar mass of the element is a key factor in reaction mechanisms. Isotopes of an element differ in the number of neutrons in the nucleus. The number of protons in the nucleus of an atom determines the particular characteristics of the element. The most stable form of the isotope therefore has a different molar mass which subsequently changes the formula weight of the compound. Differences in molar mass consequently affect the concentration of the compound in the solution thus affecting the rate of the reaction (Gomez-Gallego and Sierra 2011). <\/p>\n
Reference<\/p>\n
Go\u0301mez-Gallego, M., & Sierra, M. A. (2011). Kinetic isotope effects in the study of organometallic reaction mechanisms.\u00a0Chemical reviews,\u00a0111(8), 4857-4963.<\/p>\n","protected":false},"excerpt":{"rendered":"
Using 19F-NMR to Study the Enzyme Kinetics of Porcine Kidney Acylase I: Lab Report Name of Student Institution Affiliation USING<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-42645","post","type-post","status-publish","format-standard","hentry"],"yoast_head":"\n